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TRIGONOMETRY

MICHAEL CORRAL

Trigonometry

Michael Corral

Schoolcraft College

About the author:

Michael Corral is an Adjunct Faculty member of the Department of Mathematics at

Schoolcraft College. He received a B.A. in Mathematics from the University of California

at Berkeley, and received an M.A. in Mathematics and an M.S. in Industrial & Operations

Engineering from the University of Michigan.

This text was typeset in LATEX with the KOMA-Script bundle, using the GNU Emacs

text editor on a Fedora Linux system. The graphics were created using TikZ and Gnuplot.

Copyright © 2009 Michael Corral.

Permission is granted to copy, distribute and/or modify this document under the terms of the

GNU Free Documentation License, Version 1.3 or any later version published by the Free

Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover

Texts. A copy of the license is included in the section entitled “GNU Free Documentation

License.”

Preface

This book covers elementary trigonometry. It is suitable for a one-semester course at the

college level, though it could also be used in high schools. The prerequisites are high school

algebra and geometry.

This book basically consists of my lecture notes from teaching trigonometry at Schoolcraft

College over several years, expanded with some exercises. There are exercises at the end

of each section. I have tried to include some more challenging problems, with hints when

I felt those were needed. An average student should be able to do most of the exercises.

Answers and hints to many of the odd-numbered and some of the even-numbered exercises

are provided in Appendix A.

This text probably has a more geometric feel to it than most current trigonometry texts.

That was, in fact, one of the reasons I wanted to write this book. I think that approaching the

subject with too much of an analytic emphasis is a bit confusing to students. It makes much

of the material appear unmotivated. This book starts with the “old-fashioned” right triangle

approach to the trigonometric functions, which is more intuitive for students to grasp.

In my experience, presenting the definitions of the trigonometric functions and then im-

mediately jumping into proving identities is too much of a detour from geometry to analysis

for most students. So this book presents material in a very different order than most books

today. For example, after starting with the right triangle definitions and some applications,

general (oblique) triangles are presented. That seems like a more natural progression of

topics, instead of leaving general triangles until the end as is usually the case.

The goal of this book is a bit different, too. Instead of taking the (doomed) approach that

students have to be shown that trigonometry is “relevant to their everyday lives” (which

inevitably comes off as artificial), this book has a different mindset: preparing students

to use trigonometry as it is used in other courses. Virtually no students will ever in their

“everyday life” figure out the height of a tree with a protractor or determine the angular

speed of a Ferris wheel. Students are far more likely to need trigonometry in other courses

(e.g. engineering, physics). I think that math instructors have a duty to prepare students

for that.

In Chapter 5 students are asked to use the free open-source software Gnuplot to graph

some functions. However, any program can be used for those exercises, as long as it produces

accurate graphs. Appendix B contains a brief tutorial on Gnuplot.

There are a few exercises that require the student to write his or her own computer pro-

gram to solve some numerical computation problems. There are a few code samples in Chap-

ter 6, written in the Java and Python programming languages, hopefully sufficiently clear

so that the reader can figure out what is being done even without knowing those languages.

iii

iv

PREFACE

Octave and Sage are also mentioned. This book probably discusses numerical issues more

than most texts at this level (e.g. the numerical instability of Heron’s formula for the area

of a triangle, the secant method for solving trigonometric equations). Numerical methods

probably should have been emphasized even more in the text, since it is rare when even a

moderately complicated trigonometric equation can be solved with elementary methods, and

since mathematical software is so readily available.

I wanted to keep this book as brief as possible. Someone once joked that trigonometry

is two weeks of material spread out over a full semester, and I think that there is some

truth to that. However, some decisions had to be made on what material to leave out. I had

planned to include sections on vectors, spherical trigonometry - a subject which has basically

vanished from trigonometry texts in the last few decades (why?) - and a few other topics,

but decided against it. The hardest decision was to exclude Paul Rider’s clever geometric

proof of the Law of Tangents without using any sum-to-product identities, though I do give

a reference to it.

This book is released under the GNU Free Documentation License (GFDL), which allows

others to not only copy and distribute the book but also to modify it. For more details, see

the included copy of the GFDL. So that there is no ambiguity on this matter, anyone can

make as many copies of this book as desired and distribute it as desired, without needing

my permission. The PDF version will always be freely available to the public at no cost (go

to http://www.mecmath.net/trig). Feel free to contact me at mcorral@schoolcraft.edu for any questions on this or any other matter involving the book (e.g. comments, suggestions,

corrections, etc). I welcome your input.

July 2009

MICHAEL CORRAL

Livonia, Michigan

Contents

Preface

iii

1

Right Triangle Trigonometry

1

1.1

Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

1.2

Trigonometric Functions of an Acute Angle . . . . . . . . . . . . . . . . . . . .

7

1.3

Applications and Solving Right Triangles . . . . . . . . . . . . . . . . . . . . . 14

1.4

Trigonometric Functions of Any Angle . . . . . . . . . . . . . . . . . . . . . . . 24

1.5

Rotations and Reflections of Angles . . . . . . . . . . . . . . . . . . . . . . . . . 32

2

General Triangles

38

2.1

The Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.2

The Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.3

The Law of Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.4

The Area of a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.5

Circumscribed and Inscribed Circles . . . . . . . . . . . . . . . . . . . . . . . . 59

3

Identities

65

3.1

Basic Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.2

Sum and Difference Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.3

Double-Angle and Half-Angle Formulas . . . . . . . . . . . . . . . . . . . . . . 78

3.4

Other Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4

Radian Measure

87

4.1

Radians and Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4.2

Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

4.3

Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

4.4

Circular Motion: Linear and Angular Speed . . . . . . . . . . . . . . . . . . . . 100

5

Graphing and Inverse Functions

103

5.1

Graphing the Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . 103

5.2

Properties of Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . 109

5.3

Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 120

6

Additional Topics

129

6.1

Solving Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 129

6.2

Numerical Methods in Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . 133

v

vi

CONTENTS

6.3

Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

6.4

Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Appendix A:

Answers and Hints to Selected Exercises

152

Appendix B:

Graphing with Gnuplot

155

GNU Free Documentation License

160

History

168

Index

169

1 Right Triangle Trigonometry

Trigonometry is the study of the relations between the sides and angles of triangles. The

word “trigonometry” is derived from the Greek words trigono (τρ´ιγωνo), meaning “triangle”,

and metro (µǫτρ ´

ω), meaning “measure”. Though the ancient Greeks, such as Hipparchus

and Ptolemy, used trigonometry in their study of astronomy between roughly 150 B.C. - A.D.

200, its history is much older. For example, the Egyptian scribe Ahmes recorded some rudi-

mentary trigonometric calculations (concerning ratios of sides of pyramids) in the famous

Rhind Papyrus sometime around 1650 B.C.1

Trigonometry is distinguished from elementary geometry in part by its extensive use of

certain functions of angles, known as the trigonometric functions. Before discussing those

functions, we will review some basic terminology about angles.

1.1 Angles

Recall the following definitions from elementary geometry:

(a) An angle is acute if it is between 0◦ and 90◦.

(b) An angle is a right angle if it equals 90◦.

(c) An angle is obtuse if it is between 90◦ and 180◦.

(d) An angle is a straight angle if it equals 180◦.

(a) acute angle

(b) right angle

(c) obtuse angle

(d) straight angle

Figure 1.1.1

Types of angles

In elementary geometry, angles are always considered to be positive and not larger than

360◦. For now we will only consider such angles.2 The following definitions will be used

throughout the text:

1Ahmes claimed that he copied the papyrus from a work that may date as far back as 3000 B.C.

2Later in the text we will discuss negative angles and angles larger than 360◦.

1

2

Chapter 1 • Right Triangle Trigonometry

§1.1

(a) Two acute angles are complementary if their sum equals 90◦. In other words, if 0◦ ≤

∠ A , ∠ B ≤ 90◦ then ∠ A and ∠ B are complementary if ∠ A + ∠ B = 90◦.

(b) Two angles between 0◦ and 180◦ are supplementary if their sum equals 180◦. In other

words, if 0◦ ≤ ∠ A , ∠ B ≤ 180◦ then ∠ A and ∠ B are supplementary if ∠ A + ∠ B = 180◦.

(c) Two angles between 0◦ and 360◦ are conjugate (or explementary) if their sum equals

360◦. In other words, if 0◦ ≤ ∠ A , ∠ B ≤ 360◦ then ∠ A and ∠ B are conjugate if ∠ A+∠ B =

360◦.

∠

∠ B

B

∠ A

∠ A

∠ A

∠ B

(a) complementary

(b) supplementary

(c) conjugate

Figure 1.1.2

Types of pairs of angles

Instead of using the angle notation ∠ A to denote an angle, we will sometimes use just a

capital letter by itself (e.g. A, B, C) or a lowercase variable name (e.g. x, y, t). It is also common to use letters (either uppercase or lowercase) from the Greek alphabet, shown in

the table below, to represent angles:

Table 1.1

The Greek alphabet

Letters

Name

Letters

Name

Letters

Name

A

α

alpha

I

ι

iota

P

ρ

rho

B

β

beta

K

κ

kappa

Σ

σ

sigma

Γ

γ

gamma

Λ

λ

lambda

T

τ

tau

∆

δ

delta

M

µ

mu

Υ

υ

upsilon

E

ǫ

epsilon

N

ν

nu

Φ

φ

phi

Z

ζ

zeta

Ξ

ξ

xi

X

χ

chi

H

η

eta

O

o

omicron

Ψ

ψ

psi

Θ

θ

theta

Π

π

pi

Ω

ω

omega

In elementary geometry you learned that the sum of the angles in a triangle equals 180◦,

and that an isosceles triangle is a triangle with two sides of equal length. Recall that in a

right triangle one of the angles is a right angle. Thus, in a right triangle one of the angles

is 90◦ and the other two angles are acute angles whose sum is 90◦ (i.e. the other two angles

are complementary angles).

Angles • Section 1.1

3

Example 1.1

For each triangle below, determine the unknown angle(s):

E

Y

B

53◦

3 α

35◦

20◦

α

α

A

C

D

F

X

Z

Note: We will sometimes refer to the angles of a triangle by their vertex points. For example, in the

first triangle above we will simply refer to the angle ∠ BAC as angle A.

Solution: For triangle △ ABC, A = 35◦ and C = 20◦, and we know that A + B + C = 180◦, so 35◦ + B + 20◦ = 180◦

⇒

B = 180◦ − 35◦ − 20◦

⇒

B = 125◦ .

For the right triangle △ DEF, E = 53◦ and F = 90◦, and we know that the two acute angles D and E

are complementary, so

D + E = 90◦

⇒

D = 90◦ − 53◦

⇒

D = 37◦ .

For triangle △ X Y Z, the angles are in terms of an unknown number α, but we do know that X + Y +

Z = 180◦, which we can use to solve for α and then use that to solve for X , Y , and Z:

α + 3 α + α = 180◦

⇒

5 α = 180◦

⇒

α = 36◦

⇒

X = 36◦ , Y = 3 × 36◦ = 108◦ , Z = 36◦

Example 1.2

Thales’ Theorem states that if A, B, and C are (distinct) points on a circle such that the line segment

AB is a diameter of the circle, then the angle ∠ ACB is a right angle (see Figure 1.1.3(a)). In other words, the triangle △ ABC is a right triangle.

C

C

α β

α

β

A

B

A

B

O

O

(a)

(b)

Figure 1.1.3

Thales’ Theorem: ∠ ACB = 90◦

To prove this, let O be the center of the circle and draw the line segment OC, as in Figure 1.1.3(b).

Let α = ∠ BAC and β = ∠ ABC. Since AB is a diameter of the circle, OA and OC have the same length (namely, the circle’s radius). This means that △ OAC is an isosceles triangle, and so ∠ OC A =

∠ OAC = α. Likewise, △ OBC is an isosceles triangle and ∠ OCB = ∠ OBC = β. So we see that

∠ ACB = α+ β. And since the angles of △ ABC must add up to 180◦, we see that 180◦ = α+( α+ β)+ β =

2 ( α + β), so α + β = 90◦. Thus, ∠ ACB = 90◦.

QED

4

Chapter 1 • Right Triangle Trigonometry

§1.1

In a right triangle, the side opposite the right angle is called the hy-

B

potenuse, and the other two sides are called its legs. For example, in

c

Figure 1.1.4 the right angle is C, the hypotenuse is the line segment

a

AB, which has length c, and BC and AC are the legs, with lengths a

and b, respectively. The hypotenuse is always the longest side of a right

A

b

C

triangle (see Exercise 11).

Figure 1.1.4

By knowing the lengths of two sides of a right triangle, the length of

the third side can be determined by using the Pythagorean Theorem:

Theorem 1.1. Pythagorean Theorem: The square of the length of the hypotenuse of a

right triangle is equal to the sum of the squares of the lengths of its legs.

Thus, if a right triangle has a hypotenuse of length c and legs of lengths a and b, as in

Figure 1.1.4, then the Pythagorean Theorem says:

a 2 + b 2 = c 2

(1.1)

Let us prove this. In the right triangle △ ABC in Figure 1.1.5(a) below, if we draw a line

segment from the vertex C to the point D on the hypotenuse such that CD is perpendicular

to AB (that is, CD forms a right angle with AB), then this divides △ ABC into two smaller

triangles △ CBD and △ ACD, which are both similar to △ ABC.

B

c

d

C

D

B

a

d

b

a

c −

d

A

b

C

C

D

A

c − d

D

(a) △ ABC

(b) △ CBD

(c) △ ACD

Figure 1.1.5

Similar triangles △ ABC, △ CBD, △ ACD

Recall that triangles are similar if their corresponding angles are equal, and that similarity

implies that corresponding sides are proportional. Thus, since △ ABC is similar to △ CBD,

by proportionality of corresponding sides we see that

c

a

AB is to CB (hypotenuses) as BC is to BD (vertical legs)

⇒

=

⇒

cd = a 2 .

a

d

Since △ ABC is similar to △ ACD, comparing horizontal legs and hypotenuses gives

b

c

=

⇒

b 2 = c 2 − cd = c 2 − a 2

⇒

a 2 + b 2 = c 2 . QED

c − d

b

Note: The symbols ⊥ and ∼ denote perpendicularity and similarity, respectively. For exam-

ple, in the above proof we had CD ⊥ AB and △ ABC ∼ △ CBD ∼ △ ACD.

Angles • Section 1.1

5

Example 1.3

For each right triangle below, determine the length of the unknown side:

B

Y

E

5

z

a

2

1

1

A

4

C

D

e

F

X

1

Z

Solution: For triangle △ ABC, the Pythagorean Theorem says that

a 2 + 42 = 52

⇒

a 2 = 25 − 16 = 9

⇒

a = 3 .

For triangle △ DEF, the Pythagorean Theorem says that

e 2 + 12 = 22

⇒

e 2 = 4 − 1 = 3

⇒

e =

3 .

For triangle △ X Y Z, the Pythagorean Theorem says that

12 + 12 = z 2

⇒

z 2 = 2

⇒

z =

2 .

Example 1.4

A 17 ft ladder leaning against a wall has its foot 8 ft from the base of the wall. At

what height is the top of the ladder touching the wall?

Solution: Let h be the height at which the ladder touches the wall. We can as-

17

sume that the ground makes a right angle with the wall, as in the picture on the

h

right. Then we see that the ladder, ground, and wall form a right triangle with a

hypotenuse of length 17 ft (the length of the ladder) and legs with lengths 8 ft and

90◦

h ft. So by the Pythagorean Theorem, we have

8

h 2 + 82 = 172

⇒

h 2 = 289 − 64 = 225

⇒

h = 15 ft .

Exercises

For Exercises 1-4, find the numeric value of the indicated angle(s) for the triangle △ ABC.

1. Find B if A = 15◦ and C = 50◦.

2. Find C if A = 110◦ and B = 31◦.

3. Find A and B if C = 24◦, A = α, and B = 2 α.

4. Find A, B, and C if A = β and B = C = 4 β.

For Exercises 5-8, find the numeric value of the indicated angle(s) for the right triangle △ ABC, with

C being the right angle.

5. Find B if A = 45◦.

6. Find A and B if A = α and B = 2 α.

7. Find A and B if A = φ and B = φ 2.

8. Find A and B if A = θ and B = 1/ θ.

9. A car goes 24 miles due north then 7 miles due east. What is the straight distance between the

car’s starting point and end point?

6

Chapter 1 • Right Triangle Trigonometry

§1.1

10. One end of a rope is attached to the top of a pole 100 ft high. If the rope is 150 ft long, what is

the maximum distance along the ground from the base of the pole to where the other end can be

attached? You may assume that the pole is perpendicular to the ground.

11. Prove that the hypotenuse is the longest side in every right triangle. ( Hint: Is a 2 + b 2 > a 2 ? ) 12. Can a right triangle have sides with lengths 2, 5, and 6? Explain your answer.

13. If the lengths a, b, and c of the sides of a right triangle are positive integers, with a 2 + b 2 = c 2, then they form what is called a Pythagorean triple. The triple is normally written as ( a, b, c).

For example, (3,4,5) and (5,12,13) are well-known Pythagorean triples.

(a) Show that (6,8,10) is a Pythagorean triple.

(b) Show that if ( a, b, c) is a Pythagorean triple then so is ( ka, kb, kc) for any integer k > 0. How would you interpret this geometrically?

(c) Show that (2 mn, m 2 − n 2, m 2 + n 2) is a Pythagorean triple for all integers m > n > 0.

(d) The triple in part(c) is known as Euclid’s formula for generating Pythagorean triples. Write

down the first ten Pythagorean triples generated by this formula, i.e. use: m = 2 and n = 1;

m = 3 and n = 1, 2; m = 4 and n = 1, 2, 3; m = 5 and n = 1, 2, 3, 4.

14. This exercise will describe how to draw a line through any point outside a circle such that the

line intersects the circle at only one point. This is called a tangent line to the circle (see the picture

on the left in Figure 1.1.6), a notion which we will use throughout the text.

tangent line

A

•

O

P

C

not tangent

Figure 1.1.6

On a sheet of paper draw a circle of radius 1 inch, and call the center of that circle O. Pick a

point P which is 2.5 inches away from O. Draw the circle which has OP as a diameter, as in the

picture on the right in Figure 1.1.6. Let A be one of the points where this circle intersects the first circle. Draw the line through P and A. In general the tangent line through a point on a circle

is perpendicular to the line joining that point to the center of the circle (why?). Use this fact to

explain why the line you drew is the tangent line through A and to calculate the length of P A.

Does it match the physical measurement of P A?

15. Suppose that

C

△ ABC is a triangle with side AB a diameter of a circle

with center O, as in the picture on the right, and suppose that the

vertex C lies on the circle. Now imagine that you rotate the circle 180◦

around its center, so that △ ABC is in a new position, as indicated by

A

B

the dashed lines in the picture. Explain how this picture proves Thales’

O

Theorem.

Trigonometric Functions of an Acute Angle • Section 1.2

7

1.2 Trigonometric Functions of an Acute Angle

Consider a right triangle △ ABC, with the right angle at C and

e

B o

with lengths a, b, and c, as in the figure on the right. For the acute

pp

potenus

c

o

angle A, call the leg BC its opposite side, and call the leg AC its

hy

a sit

adjacent side. Recall that the hypotenuse of the triangle is the side

e

AB. The ratios of sides of a right triangle occur often enough in prac-

A

b

C

tical applications to warrant their own names, so we define the six

adjacent

trigonometric functions of A as follows:

Table 1.2

The six trigonometric functions of A

Name of function

Abbreviation

Definition

opposite side

a

sine A

sin A

=

=

hypotenuse

c

adjacent side

b

cosine A

cos A

=

=

hypotenuse

c

opposite side

a

tangent A

tan A

=

=

adjacent side

b

hypotenuse

c

cosecant A

csc A

=

=

opposite side

a

hypotenuse

c

secant A

sec A

=

=

adjacent side

b

adjacent side

b

cotangent A

cot A

=

=

opposite side

a

We will usually use the abbreviated names of the functions. Notice from Table 1.2 that

the pairs sin A and csc A, cos A and sec A, and tan A and cot A are reciprocals:

1

1

1

csc A =

sec A =

cot A =

sin A

cos A

tan A

1

1

1

sin A =

cos A =

tan A =

csc A

sec A

cot A

8

Chapter 1 • Right Triangle Trigonometry

§1.2

Example 1.5

For the right triangle △ ABC shown on the right, find the values of all six trigono-

B

metric functions of the acute angles A and B.

5

3

Solution: The hypotenuse of △ ABC has length 5. For angle A, the opposite side

BC has length 3 and the adjacent side AC has length 4. Thus:

A

4

C

opposite

3

adjacent

4

opposite

3

sin A =

=

cos A =

=

tan A =

=

hypotenuse

5

hypotenuse

5

adjacent

4

hypotenuse

5

hypotenuse

5

adjacent

4

csc A =

=

sec A =

=

cot A =

=

opposite

3

adjacent

4

opposite

3

For angle B, the opposite side AC has length 4 and the adjacent side BC has length 3. Thus:

opposite

4

adjacent

3

opposite

4

sin B =

=

cos B =

=

tan B =

=

hypotenuse

5

hypotenuse

5

adjacent

3

hypotenuse

5

hypotenuse

5

adjacent

3

csc B =

=

sec B =

=

cot B =

=

opposite

4

adjacent

3

opposite

4

Notice in Example 1.5 that we did not specify the units for the lengths. This raises the

possibility that our answers depended on a triangle of a specific physical size.

For example, suppose that two different students are reading this textbook: one in the

United States and one in Germany. The American student thinks that the lengths 3, 4, and

5 in Example 1.5 are measured in inches, while the German student thinks that they are

measured in centimeters. Since 1 in ≈ 2.54 cm, the students are using triangles of different

physical sizes (see Figure 1.2.1 below, not drawn to scale).

B

B

5

3

B′

B′

5

3

A

A

4

C

A′

4

C′

A′

C′

C

(a) Inches

(b) Centimeters

(c) Similar triangles

Figure 1.2.1

△ ABC ∼ △ A′ B′ C′

If the American triangle is △ ABC and the German triangle is △ A′ B′ C′, then we see

from Figure 1.2.1 that △ ABC is similar to △ A′ B′ C′, and hence the corresponding angles Trigonometric Functions of an Acute Angle • Section 1.2

9

are equal and the ratios of the corresponding sides are equal. In fact, we know that com-

mon ratio: the sides of △ ABC are approximately 2.54 times longer than the corresponding

sides of △ A′ B′ C′. So when the American student calculates sin A and the German student

calculates sin A′, they get the same answer:3

BC

AB

BC

B′ C′

△ ABC ∼ △ A′ B′ C′

⇒

=

⇒

=

⇒

sin A = sin A′

B′ C′

A′ B′

AB

A′ B′

Likewise, the other values of the trigonometric functions of A and A′ are the same. In fact,

our argument was general enough to work with any similar right triangles. This leads us to

the following conclusion:

When calculating the trigonometric functions of an acute angle A, you may

use any right triangle which has A as one of the angles.

Since we defined the trigonometric functions in terms of ratios of sides, you can think

of the units of measurement for those sides as canceling out in those ratios. This means

that the values of the trigonometric functions are unitless numbers. So when the American

student calculated 3/5 as the value of sin A in Example 1.5, that is the same as the 3/5 that

the German student calculated, despite the different units for the lengths of the sides.

Example 1.6

Find the values of all six trigonometric functions of 45◦.

1

B

Solution: Since we may use any right triangle which has 45◦ as one of the

angles, use the simplest one: take a square whose sides are all 1 unit long and

2

1

1

divide it in half diagonally, as in the figure on the right. Since the two legs

of the triangle △ ABC have the same length, △ ABC is an isosceles triangle,

45◦

which means that the angles A and B are equal. So since A + B = 90◦, this

A

1

C

means that we must have A = B = 45◦. By the Pythagorean Theorem, the

length c of the hypotenuse is given by

c 2 = 12 + 12 = 2

⇒

c =

2 .

Thus, using the angle A we get:

opposite

1

adjacent

1

opposite

1

sin 45◦ =

=

cos 45◦ =

=

tan 45◦ =

=

= 1

hypotenuse

2

hypotenuse

2

adjacent

1

hypotenuse

hypotenuse

adjacent

1

csc 45◦ =

=

2

sec 45◦ =

=

2

cot 45◦ =

=

= 1

opposite

adjacent

opposite

1

Note that we would have obtained the same answers if we had used any right triangle similar to

△ ABC. For example, if we multiply each side of △ ABC by

2, then we would have a similar

triangle with legs of length

2 and hypotenuse of length 2. This would give us sin 45◦ = 2 , which

2

equals

2

= 1 as before. The same goes for the other functions.

2· 2

2

3We will use the notation AB to denote the length of a line segment AB.

10

Chapter 1 • Right Triangle Trigonometry

§1.2

Example 1.7

Find the values of all six trigonometric functions of 60◦.

B

Solution: Since we may use any right triangle which has 60◦ as one of

the angles, we will use a simple one: take a triangle whose sides are all 2

units long and divide it in half by drawing the bisector from one vertex to

2

30◦

2

the opposite side, as in the figure on the right. Since the original triangle

3

was an equilateral triangle (i.e. all three sides had the same length), its

three angles were all the same, namely 60◦. Recall from elementary ge-

60◦

60◦

ometry that the bisector from the vertex angle of an equilateral triangle

A

1

C

1

to its opposite side bisects both the vertex angle and the opposite side. So

as in the figure on the right, the triangle

2

△ ABC has angle A = 60◦ and

angle B = 30◦, which forces the angle C to be 90◦. Thus, △ ABC is a right

triangle. We see that the hypotenuse has length c = AB = 2 and the leg AC has length b = AC = 1.

By the Pythagorean Theorem, the length a of the leg BC is given by

a 2 + b 2 = c 2

⇒

a 2 = 22 − 12 = 3

⇒

a =

3 .

Thus, using the angle A we get:

opposite

3

adjacent

1

opposite

3

sin 60◦ =

=

cos 60◦ =

=

tan 60◦ =

=

=

3

hypotenuse

2

hypotenuse

2

adjacent

1

hypotenuse

2

hypotenuse

adjacent

1

csc 60◦ =

=

sec 60◦ =

= 2

cot 60◦ =

=

opposite

3

adjacent

opposite

3

Notice that, as a bonus, we get the values of all six trigonometric functions of 30◦, by using angle

B = 30◦ in the same triangle △ ABC above:

opposite

1

adjacent

3

opposite

1

sin 30◦ =

=

cos 30◦ =

=

tan 30◦ =

=

hypotenuse

2

hypotenuse

2

adjacent

3

hypotenuse

hypotenuse

2

adjacent

3

csc 30◦ =

= 2

sec 30◦ =

=

cot 30◦ =

=

=

3

opposite

adjacent

3

opposite

1

Example 1.8

A is an acute angle such that sin A = 2 . Find the values of the other trigonometric

B

3

functions of A.

3

2

Solution: In general it helps to draw a right triangle to solve problems of this

type. The reason is that the trigonometric functions were defined in terms of

A

b

C

ratios of sides of a right triangle, and you are given one such function (the sine,

in this case) already in terms of a ratio: sin A = 2 . Since sin A is defined as

3

opposite , use 2 as the length of the side opposite A and use 3 as the length of the hypotenuse in a

hypotenuse

right triangle △ ABC (see the figure above), so that sin A = 2 . The adjacent side to A has unknown

3

length b, but we can use the Pythagorean Theorem to find it:

22 + b 2 = 32

⇒

b 2 = 9 − 4 = 5

⇒

b =

5

Trigonometric Functions of an Acute Angle • Section 1.2

11

We now know the lengths of all sides of the triangle △ ABC, so we have:

adjacent

5

opposite

2

cos A =

=

tan A =

=

hypotenuse

3

adjacent

5

hypotenuse

3

hypotenuse

3

adjacent

5

csc A =

=

sec A =

=

cot A =

=

opposite

2

adjacent

5

opposite

2

You may have noticed the connections between the sine and cosine, secant and cosecant,

and tangent and cotangent of the complementary angles in Examples 1.5 and 1.7. Generalizing those examples gives us the following theorem:

Theorem 1.2. Cofunction Theorem: If A and B are the complementary acute angles in a

right triangle △ ABC, then the following relations hold:

sin A = cos B

sec A = csc B

tan A = cot B

sin B = cos A

sec B = csc A

tan B = cot A

We say that the pairs of functions { sin, cos }, { sec, csc }, and { tan, cot } are cofunctions.

So sine and cosine are cofunctions, secant and cosecant are cofunctions, and tangent and

cotangent are cofunctions. That is how the functions cosine, cosecant, and cotangent got the

“co” in their names. The Cofunction Theorem says that any trigonometric function of an

acute angle is equal to its cofunction of the complementary angle.

Example 1.9

Write each of the following numbers as trigonometric functions of an angle less than 45◦: (a) sin 65◦;

(b) cos 78◦; (c) tan 59◦.

Solution: (a) The complement of 65◦ is 90◦ − 65◦ = 25◦ and the cofunction of sin is cos, so by the

Cofunction Theorem we know that sin 65◦ = cos 25◦.

(b) The complement of 78◦ is 90◦ − 78◦ = 12◦ and the cofunction of cos is sin, so cos 78◦ = sin 12◦.

(c) The complement of 59◦ is 90◦ − 59◦ = 31◦ and the cofunction of tan is cot, so tan 59◦ = cot 31◦.

2 a

a 2

60◦

45◦

a

a

30◦

45◦

a

a 3

(a) 45−45−90

(b) 30 − 60 − 90

Figure 1.2.2

Two general right triangles (any a > 0)

The angles 30◦, 45◦, and 60◦ arise often in applications. We can use the Pythagorean

Theorem to generalize the right triangles in Examples 1.6 and 1.7 and see what any 45 −

45 − 90 and 30 − 60 − 90 right triangles look like, as in Figure 1.2.2 above.

12

Chapter 1 • Right Triangle Trigonometry

§1.2

Example 1.10

Find the sine, cosine, and tangent of 75◦.

D

Solution: Since 75◦ = 45◦+30◦, place a 30−60−90 right triangle

△ ADB with legs of length

3 and 1 on top of the hypotenuse

of a 45 − 45 − 90 right triangle △ ABC whose hypotenuse has

1

length

3, as in the figure on the right. From Figure 1.2.2(a) we

know that the length of each leg of △ ABC is the length of the

B

hypotenuse divided by

2. So AC = BC = 3 =

3 . Draw DE

F

2

2

perpendicular to AC, so that △ ADE is a right triangle. Since

2

∠ BAC = 45◦ and ∠ D AB = 30◦, we see that ∠ D AE = 75◦ since

3

it is the sum of those two angles. Thus, we need to find the sine,

3

cosine, and tangent of ∠ D AE.

2

Notice that ∠ ADE = 15◦, since it is the complement of ∠ D AE.

And ∠ ADB = 60◦, since it is the complement of ∠ D AB. Draw

◦

30

BF perpendicular to DE, so that △ DFB is a right triangle.

Then ∠ BDF = 45◦, since it is the difference of ∠ ADB = 60◦ and

45◦

E

A

C

∠ ADE = 15◦. Also, ∠ DBF = 45◦ since it is the complement of

3

2

∠ BDF. The hypotenuse BD of △ DFB has length 1 and △ DFB

is a 45 − 45 − 90 right triangle, so we know that DF = FB = 1 .

2

Now, we know that DE ⊥ AC and BC ⊥ AC, so FE and BC are parallel. Likewise, FB and EC are both perpendicular to DE and hence FB is parallel to EC. Thus, FBCE is a rectangle, since ∠ BCE

is a right angle. So EC = FB = 1 and FE = BC =

3 . Hence,

2

2

DE = DF + FE = 1 +

3

and

AE = AC − EC =

3

=

3 − 1 . Thus,

2

2 =

3 + 1

2

2 −

1

2

2

3+1

3−1

3+1

sin 75◦ = DE

2

, cos 75◦

2

, and tan 75◦

2

.

AD =

2

= 6+ 2

4

= AE

AD =

2

= 6− 2

4

= DE

AE =

= 6+ 2

3−1

6− 2

2

Note: Taking reciprocals, we get csc 75◦ =

4

, sec 75◦ =

4

, and cot 75◦ = 6− 2 .

6+ 2

6− 2

6+ 2

Exercises

For Exercises 1-10, find the values of all six trigonometric functions of

B

angles A and B in the right triangle △ ABC in Figure 1.2.3.

c

a

1. a = 5, b = 12, c = 13

2. a = 8, b = 15, c = 17

A

b

C

3. a = 7, b = 24, c = 25

4. a = 20, b = 21, c = 29

Figure 1.2.3

5. a = 9, b = 40, c = 41

6. a = 1, b = 2, c = 5

7. a = 1, b = 3

8. a = 2, b = 5

9. a = 5, c = 6

10. b = 7, c = 8

For Exercises 11-18, find the values of the other five trigonometric functions of the acute angle A

given the indicated value of one of the functions.

Trigonometric Functions of an Acute Angle • Section 1.2

13

11. sin A = 3

12. cos A

13. cos A

14. sin A

4

= 23

= 2

= 2

10

4

15. tan A = 5

16. tan A

18. csc A

9

= 3

17. sec A = 73

= 3

For Exercises 19-23, write the given number as a trigonometric function of an acute angle less than

45◦.

19. sin 87◦

20. sin 53◦

21. cos 46◦

22. tan 66◦

23. sec 77◦

For Exercises 24-28, write the given number as a trigonometric function of an acute angle greater

than 45◦.

24. sin 1◦

25. cos 13◦

26. tan 26◦

27. cot 10◦

28. csc 43◦

29. In Example 1.7 we found the values of all six trigonometric functions of 60◦ and 30◦.

(a) Does sin 30◦ + sin 30◦ = sin 60◦?

(b) Does cos 30◦ + cos 30◦ = cos 60◦?

(c) Does tan 30◦ + tan 30◦ = tan 60◦?

(d) Does 2 sin 30◦ cos 30◦ = sin 60◦?

30. For an acute angle A, can sin A be larger than 1? Explain your answer.

31. For an acute angle A, can cos A be larger than 1? Explain your answer.

32. For an acute angle A, can sin A be larger than tan A? Explain your answer.

33. If A and B are acute angles and A < B, explain why sin A < sin B.

34. If A and B are acute angles and A < B, explain why cos A > cos B.

35. Prove the Cofunction Theorem (Theorem 1.2). ( Hint: Draw a right triangle and label the angles

and sides. )

36. Use Example 1.10 to find all six trigonometric functions of 15◦.

37. In Figure 1.2.4, CB is a diameter of a circle with a radius of

A

D

2 cm and center O, △ ABC is a right triangle, and CD

has length

3 cm.

(a) Find sin A. ( Hint: Use Thales’ Theorem. )

3

(b) Find the length of AC.

C

B

O

2

(c) Find the length of AD.

(d) Figure 1.2.4 is drawn to scale. Use a protractor to

measure the angle A, then use your calculator to find

the sine of that angle. Is the calculator result close to

your answer from part(a)?

Figure 1.2.4

Note: Make sure that your calculator is in degree mode.

38. In Exercise 37, verify that the area of △ ABC equals 1 AB

2

· CD. Why does this make sense?

39. In Exercise 37, verify that the area of △ ABC equals 1 AB

2

· AC sin A.

40. In Exercise 37, verify that the area of △ ABC equals 1 ( BC)2 cot A.

2

14

Chapter 1 • Right Triangle Trigonometry

§1.3

1.3 Applications and Solving Right Triangles

Throughout its early development, trigonometry was often used as a means of indirect mea-

surement, e.g. determining large distances or lengths by using measurements of angles and

small, known distances. Today, trigonometry is widely used in physics, astronomy, engineer-

ing, navigation, surveying, and various fields of mathematics and other disciplines. In this

section we will see some of the ways in which trigonometry can be applied. Your calculator

should be in degree mode for these examples.

Example 1.11

A person stands 150 ft away from a flagpole and measures an angle of elevation of 32◦ from his

horizontal line of sight to the top of the flagpole. Assume that the person’s eyes are a vertical distance

of 6 ft from the ground. What is the height of the flagpole?

Solution: The picture on the right describes the situation. We see that

the height of the flagpole is h + 6 ft, where

h

h

= tan 32◦

⇒

h = 150 tan 32◦ = 150 (0.6249) = 94 .

32◦

150

150

How did we know that tan 32◦ = 0.6249? By using a calculator. And

6

since none of the numbers we were given had decimal places, we rounded

off the answer for h to the nearest integer. Thus, the height of the flagpole is h +6 = 94+6 = 100 ft .

Example 1.12

A person standing 400 ft from the base of a mountain measures the angle of elevation from the ground

to the top of the mountain to be 25◦. The person then walks 500 ft straight back and measures the

angle of elevation to now be 20◦. How tall is the mountain?

Solution: We will assume that the ground is flat and not in-

clined relative to the base of the mountain. Let h be the height

of the mountain, and let x be the distance from the base of the

mountain to the point directly beneath the top of the moun-

h

tain, as in the picture on the right. Then we see that

20◦

25◦

h

= tan 25◦

⇒

h = ( x + 400) tan 25◦ , and

x

500

400

x

+ 400

h

= tan 20◦

⇒

h = ( x + 900) tan 20◦ , so

x + 400 + 500

( x + 400) tan 25◦ = ( x + 900) tan 20◦, since they both equal h. Use that equation to solve for x:

900 tan 20◦ − 400 tan 25◦

x tan 25◦ − x tan 20◦ = 900 tan 20◦ − 400 tan 25◦

⇒

x =

= 1378 ft

tan 25◦ − tan 20◦

Finally, substitute x into the first formula for h to get the height of the mountain:

h = (1378 + 400) tan 25◦ = 1778 (0.4663) = 829 ft

Applications and Solving Right Triangles • Section 1.3

15

Example 1.13

A blimp 4280 ft above the ground measures an angle of depression of 24◦ from its horizontal line of

sight to the base of a house on the ground. Assuming the ground is flat, how far away along the

ground is the house from the blimp?

Solution: Let x be the distance along the ground from the blimp

24◦

to the house, as in the picture to the right. Since the ground and the

blimp’s horizontal line of sight are parallel, we know from elementary

4280

geometry that the angle of elevation θ from the base of the house to

θ

the blimp is equal to the angle of depression from the blimp to the

base of the house, i.e. θ = 24◦. Hence,

x

4280

4280

= tan 24◦

⇒

x =

= 9613 ft .

x

tan 24◦

Example 1.14

An observer at the top of a mountain 3 miles above sea level measures an angle of depression of 2.23◦

to the ocean horizon. Use this to estimate the radius of the earth.

Solution: We will assume that the earth is a sphere.4 Let r be

A

B

the radius of the earth. Let the point A represent the top of the

3

2.23◦

mountain, and let H be the ocean horizon in the line of sight from

H

A, as in Figure 1.3.1. Let O be the center of the earth, and let B

be a point on the horizontal line of sight from A (i.e. on the line

r

perpendicular to O A). Let θ be the angle ∠ AOH.

r

θ

Since A is 3 miles above sea level, we have O A = r + 3. Also,

OH = r. Now since AB ⊥ OA, we have ∠ OAB = 90◦, so we see

O

that ∠ OAH = 90◦ − 2.23◦ = 87.77◦. We see that the line through A

and H is a tangent line to the surface of the earth (considering the

surface as the circle of radius r through H as in the picture). So

by Exercise 14 in Section 1.1, AH ⊥ OH and hence ∠ OH A = 90◦.

Since the angles in the triangle △ OAH add up to 180◦, we have

Figure 1.3.1

θ = 180◦ − 90◦ − 87.77◦ = 2.23◦. Thus,

OH

r

r

cos θ =

=

⇒

= cos 2.23◦ ,

O A

r + 3

r + 3

so solving for r we get

r = ( r + 3) cos 2.23◦

⇒

r − r cos 2.23◦ = 3 cos 2.23◦

3 cos 2.23◦

⇒

r = 1 − cos 2.23◦

⇒

r = 3958.3 miles .

Note: This answer is very close to the earth’s actual (mean) radius of 3956.6 miles.

4Of course it is not perfectly spherical. The earth is an ellipsoid, i.e. egg-shaped, with an observed ellipticity

of 1/297 (a sphere has ellipticity 0). See pp. 26-27 in W.H. MUNK AND G.J.F MACDONALD, The Rotation of the

Earth: A Geophysical Discussion, London: Cambridge University Press, 1960.

16

Chapter 1 • Right Triangle Trigonometry

§1.3

Example 1.15

As another application of trigonometry to astronomy, we will find the distance

B

from the earth to the sun. Let O be the center of the earth, let A be a point

on the equator, and let B represent an object (e.g. a star) in space, as in the

α

picture on the right. If the earth is positioned in such a way that the angle

∠ OAB = 90◦, then we say that the angle α = ∠ OBA is the equatorial parallax

of the object. The equatorial parallax of the sun has been observed to be ap-

proximately α = 0.00244◦. Use this to estimate the distance from the center of

the earth to the sun.

A

O

Solution: Let B be the position of the sun. We want to find the length of OB.

We will use the actual radius of the earth, mentioned at the end of Example

1.14, to get O A = 3956.6 miles. Since ∠ OAB = 90◦, we have

O A

O A

3956.6

= sin α

⇒

OB =

=

= 92908394 ,

OB

sin α

sin 0.00244◦

so the distance from the center of the earth to the sun is approximately 93 million miles .

Note: The earth’s orbit around the sun is an ellipse, so the actual distance to the sun varies.

In the above example we used a very small angle (0.00244◦). A degree can be divided into

smaller units: a minute is one-sixtieth of a degree, and a second is one-sixtieth of a minute.

The symbol for a minute is ′ and the symbol for a second is ′′. For example, 4.5◦ = 4◦ 30′. And

4.505◦ = 4◦ 30′ 18′′:

30

18

4◦ 30′ 18′′ = 4 +

+

degrees = 4.505◦

60

3600

In Example 1.15 we used α = 0.00244◦ ≈ 8.8′′, which we mention only because some angle

measurement devices do use minutes and seconds.

Example 1.16

An observer on earth measures an angle of 32′ 4′′ from one visible

B

edge of the sun to the other (opposite) edge, as in the picture on the

right. Use this to estimate the radius of the sun.

E

32′ 4′′

S

Solution: Let the point E be the earth and let S be the center of

the sun. The observer’s lines of sight to the visible edges of the sun

A

are tangent lines to the sun’s surface at the points A and B. Thus,

∠ E AS = ∠ EBS = 90◦. The radius of the sun equals AS. Clearly AS = BS. So since EB = E A (why?), the triangles △ E AS and △ EBS are similar. Thus, ∠ AES = ∠ BES = 1

(32′ 4′′)

2 ∠ AEB = 1

2

= 16′ 2′′ =

(16/60) + (2/3600) = 0.26722◦.

Now, ES is the distance from the surface of the earth (where the observer stands) to the cen-

ter of the sun. In Example 1.15 we found the distance from the center of the earth to the sun

to be 92, 908, 394 miles. Since we treated the sun in that example as a point, then we are justi-

fied in treating that distance as the distance between the centers of the earth and sun. So ES =

92908394 − radius of earth = 92908394 − 3956.6 = 92904437.4 miles. Hence,

AS

sin (∠ AES) =

⇒

AS = ES sin 0.26722◦ = (92904437.4) sin 0.26722◦ = 433,293 miles .

ES

Note: This answer is close to the sun’s actual (mean) radius of 432, 200 miles.

Applications and Solving Right Triangles • Section 1.3

17

You may have noticed that the solutions to the examples we have shown required at least

one right triangle. In applied problems it is not always obvious which right triangle to

use, which is why these sorts of problems can be difficult. Often no right triangle will be

immediately evident, so you will have to create one. There is no general strategy for this,

but remember that a right triangle requires a right angle, so look for places where you can

form perpendicular line segments. When the problem contains a circle, you can create right

angles by using the perpendicularity of the tangent line to the circle at a point5 with the line

that joins that point to the center of the circle. We did exactly that in Examples 1.14, 1.15,

and 1.16.

Example 1.17

The machine tool diagram on the right shows a symmetric V-block,

in which one circular roller sits on top of a smaller circular roller.

Each roller touches both slanted sides of the V-block. Find the di-

O

ameter d of the large roller, given the information in the diagram.

d

Solution: The diameter d of the large roller is twice the radius

D

OB, so we need to find OB. To do this, we will show that △ OBC

37◦

is a right triangle, then find the angle ∠ BOC, and then find BC.

B

C

The length OB will then be simple to determine.

8

P

.3

1

Since the slanted sides are tangent to each roller, ∠ OD A =

E

∠ PEC = 90◦. By symmetry, since the vertical line through the

centers of the rollers makes a 37◦ angle with each slanted side,

A

we have ∠ OAD = 37◦. Hence, since △ OD A is a right triangle,

∠ DOA is the complement of ∠ OAD. So ∠ DOA = 53◦.

Since the horizontal line segment BC is tangent to each roller, ∠ OBC = ∠ PBC = 90◦. Thus,

△ OBC is a right triangle. And since ∠ OD A = 90◦, we know that △ ODC is a right triangle. Now,

OB = OD (since they each equal the radius of the large roller), so by the Pythagorean Theorem we

have BC = DC:

BC 2 = OC 2 − OB 2 = OC 2 − OD 2 = DC 2

⇒

BC = DC

Thus, △ OBC and △ ODC are congruent triangles (which we denote by △ OBC ∼

= △ ODC), since

their corresponding sides are equal. Thus, their corresponding angles are equal. So in particular,

∠ BOC = ∠ DOC. We know that ∠ DOB = ∠ DOA = 53◦. Thus,

53◦ = ∠ DOB = ∠ BOC + ∠ DOC = ∠ BOC + ∠ BOC = 2 ∠ BOC

⇒

∠ BOC = 26.5◦ .

Likewise, since BP = EP and ∠ PBC = ∠ PEC = 90◦, △ BPC and △ EPC are congruent right triangles. Thus, BC = EC. But we know that BC = DC, and we see from the diagram that EC+ DC = 1.38.

Thus, BC + BC = 1.38 and so BC = 0.69. We now have all we need to find OB:

BC

BC

0.69

= tan ∠ BOC

⇒

OB =

=

= 1.384

OB

tan ∠ BOC

tan 26.5◦

Hence, the diameter of the large roller is d = 2 × OB = 2(1.384) = 2.768 .

5This will often be worded as the line that is tangent to the circle.

18

Chapter 1 • Right Triangle Trigonometry

§1.3

Example 1.18

A slider-crank mechanism is shown in Figure 1.3.2 below. As the piston moves downward the con-

necting rod rotates the crank in the clockwise direction, as indicated.

piston

A

a

C

conne b

c

c

ting

rod

B

θ

crank

r

O

Figure 1.3.2

Slider-crank mechanism

The point A is the center of the connecting rod’s wrist pin and only moves vertically. The point B

is the center of the crank pin and moves around a circle of radius r centered at the point O, which

is directly below A and does not move. As the crank rotates it makes an angle θ with the line O A.

The instantaneous center of rotation of the connecting rod at a given time is the point C where the

horizontal line through A intersects the extended line through O and B. From Figure 1.3.2 we see that ∠ OAC = 90◦, and we let a = AC, b = AB, and c = BC. In the exercises you will show that for 0◦ < θ < 90◦,

b 2 − r 2 (sin θ)2

c =

and

a = r sin θ +

b 2 − r 2 (sin θ)2 tan θ .

cos θ

Applications and Solving Right Triangles • Section 1.3

19

For some problems it may help to remember that when a right tri-

r

angle has a hypotenuse of length r and an acute angle θ, as in the

r sin θ

picture on the right, the adjacent side will have length r cos θ and the

θ

opposite side will have length r sin θ. You can think of those lengths

r cos θ

as the horizontal and vertical “components” of the hypotenuse.

Notice in the above right triangle that we were given two pieces of information: one of

the acute angles and the length of the hypotenuse. From this we determined the lengths of

the other two sides, and the other acute angle is just the complement of the known acute

angle. In general, a triangle has six parts: three sides and three angles. Solving a triangle

means finding the unknown parts based on the known parts. In the case of a right triangle,

one part is always known: one of the angles is 90◦.

Example 1.19

Solve the right triangle in Figure 1.3.3 using the given

B

information:

c

a

(a) c = 10, A = 22◦

Solution: The unknown parts are a, b, and B. Solving yields:

A

b

C

a = c sin A = 10 sin 22◦ = 3.75

Figure 1.3.3

b = c cos A = 10 cos 22◦ = 9.27

B = 90◦ − A = 90◦ − 22◦ = 68◦

(b) b = 8, A = 40◦

Solution: The unknown parts are a, c, and B. Solving yields:

a = tan A ⇒ a = b tan A = 8 tan 40◦ = 6.71

b

b

b

8

= cos A

⇒

c =

=

= 10.44

c

cos A

cos 40◦

B = 90◦ − A = 90◦ − 40◦ = 50◦

(c) a = 3, b = 4

Solution: The unknown parts are c, A, and B. By the Pythagorean Theorem,

c =

a 2 + b 2 =

32 + 42 =

25 = 5 .

✄

Now, tan A = a

tan−1 on your

b = 3

4 = 0.75. So how do we find A? There should be a key labeled ✂

✁

calculator, which works like this: give it a number x and it will tell you the angle θ such that

tan θ = x. In our case, we want the angle A such that tan A = 0.75:

✄

Enter: 0.75

Press: tan−1

Answer: 36.86989765

✂

✁

This tells us that A = 36.87◦, approximately. Thus B = 90◦ − A = 90◦ − 36.87◦ = 53.13◦.

✄

✄

Note: The sin−1 and cos−1 keys work similarly for sine and cosine, respectively. These keys use

✂

✁

✂

✁

the inverse trigonometric functions, which we will discuss in Chapter 5.

20

Chapter 1 • Right Triangle Trigonometry

§1.3

Exercises

1. From a position 150 ft above the ground, an observer in a build-

12◦

ing measures angles of depression of 12◦ and 34◦ to the top and

34◦

bottom, respectively, of a smaller building, as in the picture on

150

the right. Use this to find the height h of the smaller building.

h

2. Generalize Example 1.12: A person standing a ft from the base of

a mountain measures an angle of elevation α from the ground to

the top of the mountain. The person then walks b ft straight back

h

and measures an angle of elevation β to the top of the mountain,

β

α

as in the picture on the right. Assuming the ground is level, find

b

a

a formula for the height h of the mountain in terms of a, b, α,

and β.

3. As the angle of elevation from the top of a tower to the sun decreases from 64◦ to 49◦ during the

day, the length of the shadow of the tower increases by 92 ft along the ground. Assuming the

ground is level, find the height of the tower.

4. Two banks of a river are parallel, and the distance between two

A

500

B

points A and B along one bank is 500 ft. For a point C on the

56◦

41◦

opposite bank, ∠ BAC = 56◦ and ∠ ABC = 41◦, as in the picture

w

on the right. What is the width w of the river?

( Hint: Divide AB into two pieces. )

C

5. A tower on one side of a river is directly east and north of

N

E

points A and B, respectively, on the other side of the river.

h

The top of the tower has angles of elevation α and β from A

and B, respectively, as in the picture on the right. Let d be

the distance between A and B. Assuming that both sides of

β

the river are at the same elevation, show that the height h of

α

the tower is

d

h =

.

(cot α)2 + (cot β)2

B

A

d

6. The equatorial parallax of the moon has been observed to be approximately 57′. Taking the radius

of the earth to be 3956.6 miles, estimate the distance from the center of the earth to the moon.

( Hint: See Example 1.15. )

7. An observer on earth measures an angle of 31′ 7′′ from one visible edge of the moon to the other

(opposite) edge. Use this to estimate the radius of the moon. ( Hint: Use Exercise 6 and see Example

1.16. )

Applications and Solving Right Triangles • Section 1.3

21

8. A ball bearing sits between two metal grooves, with the top groove having an

angle of 120◦ and the bottom groove having an angle of 90◦, as in the picture

1 ′′

120◦

2

on the right. What must the diameter of the ball bearing be for the distance

90◦

between the vertexes of the grooves to be half an inch? You may assume that

the top vertex is directly above the bottom vertex.

1.7

9. The machine tool diagram on the right shows a symmetric worm

thread, in which a circular roller of diameter 1.5 inches sits.

d

Find the amount d that the top of the roller rises above the

top of the thread, given the information in the diagram. ( Hint:

Extend the slanted sides of the thread until they meet at a point. )

1.5

10. Repeat Exercise 9 using 1.8 inches as the distance across the

top of the worm thread.

11. In Exercise 9, what would the distance across the top of the

30◦

worm thread have to be to make d equal to 0 inches?

12. For 0◦ < θ < 90◦ in the slider-crank mechanism in Example 1.18, show that

b 2 − r 2 (sin θ)2

c =

and

a = r sin θ +

b 2 − r 2 (sin θ)2 tan θ .

cos θ

( Hint: In Figure 1.3.2 draw line segments from B perpendicular to O A and AC. )

′′

13. The machine tool diagram on the right shows a symmetric die punch.

1 12

In this view, the rounded tip is part of a circle of radius r, and the slanted

54◦

r

sides are tangent to that circle and form an angle of 54◦. The top and

′′

2 18

bottom sides of the die punch are horizontal. Use the information in the

diagram to find the radius r.

B

14. In the figure on the right, ∠ BAC = θ and BC = a. Use this to find

a

AB, AC, AD, DC, CE, and DE in terms of θ and a.

( Hint: What is the angle ∠ ACD ? )

θ

A

C

D

E

For Exercises 15-23, solve the right triangle in Figure 1.3.4 using the

B

given information.

c

a

15. a = 5, b = 12

16. c = 6, B = 35◦

17. b = 2, A = 8◦

A

b

C

18. a = 2, c = 7

19. a = 3, A = 26◦

20. b = 1, c = 2

Figure 1.3.4

21. b = 3, B = 26◦

22. a = 2, B = 8◦

23. c = 2, A = 45◦

22

Chapter 1 • Right Triangle Trigonometry

§1.3

24. In Example 1.10 in Section 1.2, we found the exact values of all six trigonometric functions of

75◦. For example, we showed that cot 75◦ = 6− 2 . So since tan 15◦ = cot 75◦ by the Cofunction

6+ 2

Theorem, this means that tan 15◦ = 6− 2 . We will now describe another method for finding the

6+ 2

exact values of the trigonometric functions of 15◦. In fact, it can be used to find the exact values

for the trigonometric functions of θ when those for θ are known, for any 0◦

2

< θ < 90◦. The method

is illustrated in Figure 1.3.5 and is described below.

P

60◦

30◦

15◦

7.5◦

C

B

A

O

Q

1

Figure 1.3.5

Draw a semicircle of radius 1 centered at a point O on a horizontal line. Let P be the point on the

semicircle such that OP makes an angle of 60◦ with the horizontal line, as in Figure 1.3.5. Draw

a line straight down from P to the horizontal line at the point Q. Now create a second semicircle

as follows: Let A be the left endpoint of the first semicircle, then draw a new semicircle centered

at A with radius equal to AP. Then create a third semicircle in the same way: Let B be the left

endpoint of the second semicircle, then draw a new semicircle centered at B with radius equal to

BP.

This procedure can be continued indefinitely to create more semicircles. In general, it can be

shown that the line segment from the center of the new semicircle to P makes an angle with the

horizontal line equal to half the angle from the previous semicircle’s center to P.

(a) Explain why ∠ P AQ = 30◦. ( Hint: What is the supplement of 60◦ ? )

(b) Explain why ∠ PBQ = 15◦ and ∠ PCQ = 7.5◦.

(c) Use Figure 1.3.5 to find the exact values of sin 15◦, cos 15◦, and tan 15◦. ( Hint: To start, you

will need to use ∠ POQ = 60◦ and OP = 1 to find the exact lengths of PQ and OQ. )

(d) Use Figure 1.3.5 to calculate the exact value of tan 7.5◦.

(e) Use the same method but with an initial angle of ∠ POQ = 45◦ to find the exact values of

sin 22.5◦, cos 22.5◦, and tan 22.5◦.

Applications and Solving Right Triangles • Section 1.3

23

25. A manufacturer needs to place ten identical ball bearings against

r

the inner side of a circular container such that each ball bear-

ing touches two other ball bearings, as in the picture on the

right. The (inner) radius of the container is 4 cm.

4

(a) Find the common radius r of the ball bearings.

(b) The manufacturer needs to place a circular ring

inside the container. What is the largest possible

(outer) radius of the ring such that it is not on top

of the ball bearings and its base is level with the

base of the container?

26. A circle of radius 1 is inscribed inside a polygon with eight sides

of equal length, called a regular octagon. That is, each of the eight

sides is tangent to the circle, as in the picture on the right.

1

(a) Calculate the area of the octagon.

(b) If you were to increase the number of sides of the

polygon, would the area inside it increase or decrease?

What number would the area approach, if any? Explain.

(c) Inscribe a regular octagon inside the same circle. That is,

draw a regular octagon such that each of its eight vertexes

touches the circle. Calculate the area of this octagon.

27. The picture on the right shows a cube whose sides are of length

A

a > 0.

(a) Find the length of the diagonal line segment AB.

(b) Find the angle θ that AB makes with the base of the cube.

θ

B

B

28. In Figure 1.3.6, suppose that α, β, and AD are known. Show that:

AD

(a) BC = cot α−cot β

AD · tan β

(b) AC = tan β−tan α

α

β

AD · sin α

A

D

C

(c) BD = sin ( β− α)

Figure 1.3.6

( Hint: What is the measure of the angle ∠ ABD ? )

29. Persons A and B are at the beach, their eyes are 5 ft and 6 ft, respectively, above sea level. How

many miles farther out is Person B’s horizon than Person A’s? (Note: 1 mile = 5280 ft)

24

Chapter 1 • Right Triangle Trigonometry

§1.4

1.4 Trigonometric Functions of Any Angle

To define the trigonometric functions of any angle - including angles less than 0◦ or greater

than 360◦ - we need a more general definition of an angle. We say that an angle is formed

−−→

by rotating a ray O A about the endpoint O (called the vertex), so that the ray is in a new

−−→

−−→

−−→

position, denoted by the ray OB. The ray O A is called the initial side of the angle, and OB

is the terminal side of the angle (see Figure 1.4.1(a)).

B

B

counter-clockwise

direction (+)

l side

ina

term

clockwise

A

O

A

direction (−)

O

initial side

(a) angle ∠ AOB

(b) positive and negative angles

Figure 1.4.1

Definition of a general angle

We denote the angle formed by this rotation as ∠ AOB, or simply ∠ O, or even just O. If

the rotation is counter-clockwise then we say that the angle is positive, and the angle is

negative if the rotation is clockwise (see Figure 1.4.1(b)).

−−→

One full counter-clockwise rotation of O A back onto itself (called a revolution), so that

the terminal side coincides with the initial side, is an angle of 360◦; in the clockwise direction

−−→

this would be −360◦.6 Not rotating OA constitutes an angle of 0◦. More than one full rotation

creates an angle greater than 360◦. For example, notice that 30◦ and 390◦ have the same

terminal side in Figure 1.4.2, since 30 + 360 = 390.

30◦

390◦

Figure 1.4.2

Angle greater than 360◦

6The system of measuring angles in degrees, such that 360◦ is one revolution, originated in ancient Babylonia.

It is often assumed that the number 360 was used because the Babylonians (supposedly) thought that there

were 360 days in a year (a year, of course, is one full revolution of the Earth around the Sun). However, there is

another, perhaps more likely, explanation which says that in ancient times a person could travel 12 Babylonian

miles in one day (i.e. one full rotation of the Earth about its axis). The Babylonian mile was large enough

(approximately 7 of our miles) to be divided into 30 equal parts for convenience, thus giving 12 × 30 = 360 equal

parts in a full rotation. See p.26 in H. EVES, An Introduction to the History of Mathematics, 5th ed., New York:

Saunders College Publishing, 1983.

Trigonometric Functions of Any Angle • Section 1.4

25

We can now define the trigonometric functions of any angle in terms of Cartesian coor-

dinates. Recall that the x y-coordinate plane consists of points denoted by pairs ( x, y) of

real numbers. The first number, x, is the point’s x coordinate, and the second number, y, is

its y coordinate. The x and y coordinates are measured by their positions along the x-axis and y-axis, respectively, which determine the point’s position in the plane. This divides the

x y-coordinate plane into four quadrants (denoted by QI, QII, QIII, QIV), based on the signs

of x and y (see Figure 1.4.3(a)-(b)).

y

y

(2, 3)

QII

QI

x < 0

x > 0

(−3,2)

y > 0

y > 0

x

x

0

0

QIII

QIV

x < 0

x > 0

(−2,−2)

y < 0

y < 0

(3, −3)

(a) Quadrants I-IV

(b) Points in the plane

y

( x, y)

r

θ

x

0

(c) Angle θ in the plane

Figure 1.4.3

x y-coordinate plane

Now let θ be any angle. We say that θ is in standard position if its initial side is the

positive x-axis and its vertex is the origin (0, 0). Pick any point ( x, y) on the terminal side of

θ a distance r > 0 from the origin (see Figure 1.4.3(c)). (Note that r =

x 2 + y 2. Why?) We

then define the trigonometric functions of θ as follows:

y

x

y

sin θ =

cos θ =

tan θ =

(1.2)

r

r

x

r

r

x

csc θ =

sec θ =

cot θ =

(1.3)

y

x

y

As in the acute case, by the use of similar triangles these definitions are well-defined (i.e.

they do not depend on which point ( x, y) we choose on the terminal side of θ). Also, notice

that |sin θ| ≤ 1 and |cos θ| ≤ 1, since | y| ≤ r and | x| ≤ r in the above definitions.

26

Chapter 1 • Right Triangle Trigonometry

§1.4

Notice that in the case of an acute angle these definitions are equivalent to our earlier

definitions in terms of right triangles: draw a right triangle with angle θ such that x =

adjacent side, y = opposite side, and r = hypotenuse. For example, this would give us sin θ =

y

opposite

and cos θ

adjacent , just as before (see Figure 1.4.4(a)).

r = hypotenuse

= xr = hypotenuse

y

y

( x, y)

e

90◦

QII

QI

r

90◦ < θ < 180◦

0◦

y

< θ < 90◦

potenus

hy

opposite side

180◦

0◦

x

0

θ

x

QIII

QIV

0

180◦

270◦

x

< θ < 270◦

< θ < 360◦

adjacent side

270◦

(a) Acute angle θ

(b) Angles by quadrant

Figure 1.4.4

In Figure 1.4.4(b) we see in which quadrants or on which axes the terminal side of an angle

0◦ ≤ θ < 360◦ may fall. From Figure 1.4.3(a) and formulas (1.2) and (1.3), we see that we can get negative values for a trigonometric function. For example, sin θ < 0 when y < 0. Figure

1.4.5 summarizes the signs (positive or negative) for the trigonometric functions based on

the angle’s quadrant:

y

QII

QI

sin +

sin +

cos −

cos +

tan −

tan +

csc +

csc +

sec −

sec +

cot −

cot +

x

0

QIII

QIV

sin −

sin −

cos −

cos +

tan +

tan −

csc −

csc −

sec −

sec +

cot +

cot −

Figure 1.4.5

Signs of the trigonometric functions by quadrant

Trigonometric Functions of Any Angle • Section 1.4

27

Example 1.20

Find the exact values of all six trigonometric functions of 120◦.

y

(−1, 3)

Solution: We know 120◦ = 180◦ − 60◦. By Example 1.7 in Section 1.2,

we see that we can use the point (−1, 3) on the terminal side of the

2

angle 120◦ in QII, since we saw in that example that a basic right

3

120◦

triangle with a 60◦ angle has adjacent side of length 1, opposite side

of length

3, and hypotenuse of length 2, as in the figure on the right.

60◦

x

Drawing that triangle in QII so that the hypotenuse is on the terminal

0

side of 120◦ makes r = 2, x = −1, and y = 3. Hence:

1

y

3

x

−1

y

3

sin 120◦ =

=

cos 120◦ =

=

tan 120◦ =

=

= − 3

r

2

r

2

x

−1

r

2

r

2

x

−1

csc 120◦ =

=

sec 120◦ =

=

= −2

cot 120◦ =

=

y

3

x

−1

y

3

Example 1.21

Find the exact values of all six trigonometric functions of 225◦.

y

Solution: We know that 225◦ = 180◦ + 45◦. By Example 1.6 in Section

225◦

1

1.2, we see that we can use the point (−1,−1) on the terminal side of

x

the angle 225◦ in QIII, since we saw in that example that a basic right

0

triangle with a 45◦ angle has adjacent side of length 1, opposite side

45◦

of length 1, and hypotenuse of length

2, as in the figure on the right.

1

Drawing that triangle in QIII so that the hypotenuse is on the terminal

2

side of 225◦ makes r = 2, x = −1, and y = −1. Hence:

(−1,−1)

y

−1

x

−1

y

−1

sin 225◦ =

=

cos 225◦ =

=

tan 225◦ =

=

= 1

r

2

r

2

x

−1

r

r

x

−1

csc 225◦ =

= − 2

sec 225◦ =

= − 2

cot 225◦ =

=

= 1

y

x

y

−1

Example 1.22

Find the exact values of all six trigonometric functions of 330◦.

y

Solution: We know that 330◦ = 360◦−30◦. By Example 1.7 in Section

3

1.2, we see that we can use the point ( 3, −1) on the terminal side of

x

the angle 225◦ in QIV, since we saw in that example that a basic right

0

30◦

triangle with a 30◦ angle has adjacent side of length

3, opposite

330◦

1

side of length 1, and hypotenuse of length 2, as in the figure on the

2

right. Drawing that triangle in QIV so that the hypotenuse is on the

( 3, −1)

terminal side of 330◦ makes r = 2, x = 3, and y = −1. Hence:

y

−1

x

3

y

−1

sin 330◦ =

=

cos 330◦ =

=

tan 330◦ =

=

r

2

r

2

x

3

r

r

2

x

csc 330◦ =

= −2

sec 330◦ =

=

cot 330◦ =

= − 3

y

x

3

y

28

Chapter 1 • Right Triangle Trigonometry

§1.4

Example 1.23

Find the exact values of all six trigonometric functions of 0◦, 90◦,

y

180◦, and 270◦.

90◦

(0, 1)

Solution: These angles are different from the angles we have con-

sidered so far, in that the terminal sides lie along either the x-axis

or the y-axis. So unlike the previous examples, we do not have any

180◦

0◦

right triangles to draw. However, the values of the trigonometric

x

(

0

functions are easy to calculate by picking the simplest points on their

−1,0)

(1, 0)

terminal sides and then using the definitions in formulas (1.2) and

(1.3).

270◦

(0, −1)

For instance, for the angle 0◦ use the point (1, 0) on its terminal

side (the positive x-axis), as in Figure 1.4.6. You could think of the

Figure 1.4.6

line segment from the origin to the point (1, 0) as sort of a degenerate

right triangle whose height is 0 and whose hypotenuse and base have

the same length 1. Regardless, in the formulas we would use r = 1, x = 1, and y = 0. Hence:

y

0

x

1

y

0

sin 0◦ =

=

= 0

cos 0◦ =

=

= 1

tan 0◦ =

=

= 0

r

1

r

1

x

1

r

1

r

1

x

1

csc 0◦ =

=

= undefined

sec 0◦ =

=

= 1

cot 0◦ =

=

= undefined

y

0

x

1

y

0

Note that csc 0◦ and cot 0◦ are undefined, since division by 0 is not allowed.

Similarly, from Figure 1.4.6 we see that for 90◦ the terminal side is the positive y-axis, so use the

point (0, 1). Again, you could think of the line segment from the origin to (0, 1) as a degenerate right

triangle whose base has length 0 and whose height equals the length of the hypotenuse. We have

r = 1, x = 0, and y = 1, and hence:

y

1

x

0

y

1

sin 90◦ =

=

= 1

cos 90◦ =

=

= 0

tan 90◦ =

=

= undefined

r

1

r

1

x

0

r

1

r

1

x

0

csc 90◦ =

=

= 1

sec 90◦ =

=

= undefined

cot 90◦ =

=

= 0

y

1

x

0

y

1

Likewise, for 180◦ use the point (−1,0) so that r = 1, x = −1, and y = 0. Hence:

y

0

x

−1

y

0

sin 180◦ =

=

= 0

cos 180◦ =

=

= −1

tan 180◦ =

=

= 0

r

1

r

1

x

−1

r

1

r

1

x

−1

csc 180◦ =

=

= undefined

sec 180◦ =

=

= −1

cot 180◦ =

=

= undefined

y

0

x

−1

y

0

Lastly, for 270◦ use the point (0, −1) so that r = 1, x = 0, and y = −1. Hence:

y

−1

x

0

y

−1

sin 270◦ =

=

= −1

cos 270◦ =

=

= 0

tan 270◦ =

=

= undefined

r

1

r

1

x

0

r

1

r

1

x

0

csc 270◦ =

=

= −1

sec 270◦ =

=

= undefined

cot 270◦ =

=

= 0

y

−1

x

0

y

−1

Trigonometric Functions of Any Angle • Section 1.4

29

The following table summarizes the values of the trigonometric functions of angles be-

tween 0◦ and 360◦ which are integer multiples of 30◦ or 45◦:

Table 1.3

Table of trigonometric function values

Angle

sin

cos

tan

csc

sec

cot

0◦

0

1

0

undefined

1

undefined

30◦

1

3

1

2

2

3

2

2

3

3

45◦

1

1

1

2

2

1

2

2

60◦

3

1

3

2

2

1

2

2

3

3

90◦

1

0

undefined

1

undefined

0

120◦

3

2

− 12

− 3

2

−2

− 1

3

3

135◦

1

− 1

−1

2

− 2

−1

2

2

150◦

1

2

2

− 3

2

− 1

− 2

− 3

3

3

180◦

0

−1

0

undefined

−1

undefined

210◦

− 1

1

3

2

− 3

2

−2

− 2

3

3

225◦

− 1

− 1

1

− 2

− 2

1

2

2

240◦

− 3

3

2

− 12

− 2

−2

1

3

3

270◦

−1

0

undefined

−1

undefined

0

300◦

− 3

1

2

2

2

− 3

− 2

− 1

3

3

315◦

− 1

1

−1

− 2

2

−1

2

2

330◦

− 1

3

2

2

− 1

−2

2

− 3

3

3

Since 360◦ represents one full revolution, the trigonometric function values repeat every

360◦. For example, sin 360◦ = sin 0◦, cos 390◦ = cos 30◦, tan 540◦ = tan 180◦, sin (−45◦) =

sin 315◦, etc. In general, if two angles differ by an integer multiple of 360◦ then each trigono-

metric function will have equal values at both angles. Angles such as these, which have the

same initial and terminal sides, are called coterminal.

In Examples 1.20-1.22, we saw how the values of trigonometric functions of an angle θ

larger than 90◦ were found by using a certain acute angle as part of a right triangle. That

acute angle has a special name: if θ is a nonacute angle then we say that the reference

angle for θ is the acute angle formed by the terminal side of θ and either the positive or

30

Chapter 1 • Right Triangle Trigonometry

§1.4

negative x-axis. So in Example 1.20, we see that 60◦ is the reference angle for the nonacute

angle θ = 120◦; in Example 1.21, 45◦ is the reference angle for θ = 225◦; and in Example

1.22, 30◦ is the reference angle for θ = 330◦.

Example 1.24

Let θ

y

= 928◦.

(a) Which angle between 0◦ and 360◦ has the same terminal side

(and hence the same trigonometric function values) as θ ?

208◦

(b) What is the reference angle for θ ?

x

0

Solution: (a) Since 928◦ = 2 × 360◦ + 208◦, then θ has the same ter-

28◦

minal side as 208◦, as in Figure 1.4.7.

928◦

(b) 928◦ and 208◦ have the same terminal side in QIII, so the refer-

ence angle for θ = 928◦ is 208◦ − 180◦ = 28◦.

Figure 1.4.7

Example 1.25

Suppose that cos θ = − 4 . Find the exact values of sin θ and tan θ.

5

Solution: We can use a method similar to the one used to solve Example 1.8 in Section 1.2. That is,

draw a right triangle and interpret cos θ as the ratio adjacent of two of its sides. Since cos θ

, we

hypotenuse

= − 45

can use 4 as the length of the adjacent side and 5 as the length of the hypotenuse. By the Pythagorean

Theorem, the length of the opposite side must then be 3. Since cos θ is negative, we know from Figure

1.4.5 that θ must be in either QII or QIII. Thus, we have two possibilities, as shown in Figure 1.4.8

below:

y

y

(−4,3)

θ

4

5

x

3

θ

0

3

x

0

5

4

(−4,−3)

(a) θ in QII

(b) θ in QIII

Figure 1.4.8

cos θ = − 45

When θ is in QII, we see from Figure 1.4.8(a) that the point (−4,3) is on the terminal side of θ, and so we have x

y

y

= −4, y = 3, and r = 5. Thus, sin θ =

and tan θ

.

r = 3

5

= x = 3

−4

When θ is in QIII, we see from Figure 1.4.8(b) that the point (−4,−3) is on the terminal side of θ, and so we have x

y

y

= −4, y = −3, and r = 5. Thus, sin θ =

and tan θ

.

r = −3

5

= x = −3

−4 = 3

4

Thus, either sin θ = 3 and tan θ

or sin θ

and tan θ

.

5

= − 34

= − 35

= 34

Since reciprocals have the same sign, csc θ and sin θ have the same sign, sec θ and cos θ

have the same sign, and cot θ and tan θ have the same sign. So it suffices to remember the

signs of sin θ, cos θ, and tan θ:

Trigonometric Functions of Any Angle • Section 1.4

31

For an angle θ in standard position and a point ( x, y) on its terminal side:

(a) sin θ has the same sign as y

(b) cos θ has the same sign as x

(c) tan θ is positive when x and y have the same sign

(d) tan θ is negative when x and y have opposite signs

Exercises

For Exercises 1-10, state in which quadrant or on which axis the given angle lies.

1. 127◦

2. −127◦

3. 313◦

4. −313◦

5. −90◦

6. 621◦

7. 230◦

8. 2009◦

9. 1079◦

10. −514◦

11. In which quadrant(s) do sine and cosine have the same sign?

12. In which quadrant(s) do sine and cosine have the opposite sign?

13. In which quadrant(s) do sine and tangent have the same sign?

14. In which quadrant(s) do sine and tangent have the opposite sign?

15. In which quadrant(s) do cosine and tangent have the same sign?

16. In which quadrant(s) do cosine and tangent have the opposite sign?

For Exercises 17-21, find the reference angle for the given angle.

17. 317◦

18. 63◦

19. −126◦

20. 696◦

21. 275◦

For Exercises 22-26, find the exact values of sin θ and tan θ when cos θ has the indicated value.

22. cos θ = 1

23. cos θ

24. cos θ

26. cos θ

2

= − 12

= 0

25. cos θ = 25

= 1

For Exercises 27-31, find the exact values of cos θ and tan θ when sin θ has the indicated value.

27. sin θ = 1

28. sin θ

29. sin θ

31. sin θ

2

= − 12

= 0

30. sin θ = − 23

= 1

For Exercises 32-36, find the exact values of sin θ and cos θ when tan θ has the indicated value.

32. tan θ = 1

33. tan θ

34. tan θ

36. tan θ

2

= − 12

= 0

35. tan θ = 5

12

= 1

For Exercises 37-40, use Table 1.3 to answer the following questions.

37. Does sin 180◦ + sin 45◦ = sin 225◦ ?

38. Does tan 300◦ − tan 30◦ = tan 270◦ ?

39. Does cos 180◦ − cos 60◦ = cos 120◦ ?

40. Does cos 240◦ = (cos 120◦)2 − (sin 120◦)2 ?

41. Expand Table 1.3 to include all integer multiples of 15◦. See Example 1.10 in Section 1.2.

32

Chapter 1 • Right Triangle Trigonometry

§1.5

1.5 Rotations and Reflections of Angles

Now that we know how to deal with angles of any measure, we will take a look at how certain

geometric operations can help simplify the use of trigonometric functions of any angle, and

how some basic relations between those functions can be made. The two operations on which

we will concentrate in this section are rotation and reflection.

To rotate an angle means to rotate its terminal side around the origin when the angle

is in standard position. For example, suppose we rotate an angle θ around the origin by 90◦

in the counterclockwise direction. In Figure 1.5.1 we see an angle θ in QI which is rotated

by 90◦, resulting in the angle θ + 90◦ in QII. Notice that the complement of θ in the right

triangle in QI is the same as the supplement of the angle θ + 90◦ in QII, since the sum of θ,

its complement, and 90◦ equals 180◦. This forces the other angle of the right triangle in QII

to be θ.

y

( x, y)

(− y, x)

r

r

θ + 90◦

θ

y

x

90◦

θ

x

y

x

Figure 1.5.1

Rotation of an angle θ by 90◦

Thus, the right triangle in QI is similar to the right triangle in QII, since the triangles

have the same angles. The rotation of θ by 90◦ does not change the length r of its terminal

side, so the hypotenuses of the similar right triangles are equal, and hence by similarity

the remaining corresponding sides are also equal. Using Figure 1.5.1 to match up those

corresponding sides shows that the point (− y, x) is on the terminal side of θ +90◦ when ( x, y)

is on the terminal side of θ. Hence, by definition,

x

− y

x

sin ( θ + 90◦) =

= cos θ , cos ( θ + 90◦) =

= −sin θ , tan ( θ + 90◦) =

= −cot θ .

r

r

− y

Though we showed this for θ in QI, it is easy (see Exercise 4) to use similar arguments for

the other quadrants. In general, the following relations hold for all angles θ:

sin ( θ + 90◦) = cos θ

(1.4)

cos ( θ + 90◦) = −sin θ

(1.5)

tan ( θ + 90◦) = −cot θ

(1.6)

Rotations and Reflections of Angles • Section 1.5

33

Example 1.26

Recall that any nonvertical line in the x y-coordinate plane can be written as y = mx + b, where m is the slope of the line (defined as m = rise ) and b is the y-intercept, i.e. where the line crosses the y-

run

axis (see Figure 1.5.2(a)). We will show that the slopes of perpendicular lines are negative reciprocals.

That is, if y = m 1 x + b 1 and y = m 2 x + b 2 are nonvertical and nonhorizontal perpendicular lines, then m 2 = − 1 (see Figure 1.5.2(b)).

m 1

y

y = mx + b

y

y = m 2 x + b 2

y = m 1 x + b 1

rise

b 2

b

0

run

m = rise

run

x

x

0

b 1

(a) Slope of a line

(b) Perpendicular lines

Figure 1.5.2

First, suppose that a line y = mx + b has nonzero slope. The line crosses the x-axis somewhere,

so let θ be the angle that the positive x-axis makes with the part of the line above the x-axis, as in

Figure 1.5.3. For m > 0 we see that θ is acute and tan θ = rise

run = m.

y

y

y

y = mx + b

= mx + b

rise

rise > 0

< 0

θ

0

θ

x

x

0

run > 0

b

run > 0

b

(a) m > 0, θ acute, rise > 0

(b) m < 0, θ obtuse, rise < 0

Figure 1.5.3

If m < 0, then we see that θ is obtuse and the rise is negative. Since the run is always positive,

our definition of tan θ from Section 1.4 means that tan θ = −rise

−run = rise

run = m (just imagine in Figure

1.5.3(b) the entire line being shifted horizontally to go through the origin, so that θ is unchanged and the point (−run,−rise) is on the terminal side of θ). Hence:

For a line y = mx + b with m = 0, the slope is given by m = tan θ , where θ is the angle formed by the positive x-axis and the part of the line above the x-axis.

Now, in Figure 1.5.2(b) we see that if two lines y = m 1 x + b 1 and y = m 2 x + b 2 are perpendicular then rotating one line counterclockwise by 90◦ around the point of intersection gives us the second

line. So if θ is the angle that the line y = m 1 x + b 1 makes with the positive x-axis, then θ + 90◦ is the angle that the line y = m 2 x + b 2 makes with the positive x-axis. So by what we just showed,

m 1 = tan θ and m 2 = tan ( θ + 90◦). But by formula (1.6) we know that tan ( θ + 90◦) = −cot θ. Hence, m 2 = −cot θ = − 1

.

QED

tan θ = − 1

m 1

34

Chapter 1 • Right Triangle Trigonometry

§1.5

Rotating an angle θ by 90◦ in the clockwise direction results in the angle θ −90◦. We could

use another geometric argument to derive trigonometric relations involving θ −90◦, but it is

easier to use a simple trick: since formulas (1.4)−(1.6) hold for any angle θ, just replace θ by θ − 90◦ in each formula. Since ( θ − 90◦) + 90◦ = θ, this gives us:

sin ( θ − 90◦) = −cos θ

(1.7)

cos ( θ − 90◦) = sin θ

(1.8)

tan ( θ − 90◦) = −cot θ

(1.9)

We now consider rotating an angle θ by 180◦. Notice from Figure 1.5.4 that the angles

θ ± 180◦ have the same terminal side, and are in the quadrant opposite θ.

y

y

( x, y)

(− x,− y)

θ + 180◦

r

θ + 180◦

r

180◦

180◦

θ

x

x

−180◦

θ

r

−180◦

r

(− x,− y)

θ − 180◦

θ − 180◦

( x, y)

(a) QI and QIII

(b) QII and QIV

Figure 1.5.4

Rotation of θ by ±180◦

Since (− x,− y) is on the terminal side of θ ± 180◦ when ( x, y) is on the terminal side of θ,

we get the following relations, which hold for all θ:

sin ( θ ± 180◦) = −sin θ

(1.10)

cos ( θ ± 180◦) = −cos θ

(1.11)

tan ( θ ± 180◦) = tan θ

(1.12)

A reflection is simply the mirror image of an object. For example, in

y

Figure 1.5.5 the original object is in QI, its reflection around the y-axis

is in QII, and its reflection around the x-axis is in QIV. Notice that if we

c

c

first reflect the object in QI around the y-axis and then follow that with

c

c x

a reflection around the x-axis, we get an image in QIII. That image is the

reflection around the origin of the original object, and it is equivalent to a

rotation of 180◦ around the origin. Notice also that a reflection around the

Figure 1.5.5

y-axis is equivalent to a reflection around the x-axis followed by a rotation

of 180◦ around the origin.

Rotations and Reflections of Angles • Section 1.5

35

Applying this to angles, we see that the reflection of an angle θ around the x-axis is the

angle − θ, as in Figure 1.5.6.

y

y

( x, y)

( x, y)

r

r

θ

θ

x

x

− θ

r

r

− θ

( x, − y)

( x, − y)

(a) QI and QIV

(b) QII and QIII

Figure 1.5.6

Reflection of θ around the x-axis

So we see that reflecting a point ( x, y) around the x-axis just replaces y by − y. Hence:

sin (− θ) = −sin θ

(1.13)

cos (− θ) = cos θ

(1.14)

tan (− θ) = −tan θ

(1.15)

Notice that the cosine function does not change in formula (1.14) because it depends on x,

and not on y, for a point ( x, y) on the terminal side of θ.

In general, a function f ( x) is an even function if f (− x) = f ( x) for all x, and it is called an odd function if f (− x) = − f ( x) for all x. Thus, the cosine function is even, while the sine and tangent functions are odd.

Replacing θ by − θ in formulas (1.4)−(1.6), then using formulas (1.13)−(1.15), gives: sin (90◦ − θ) = cos θ

(1.16)

cos (90◦ − θ) = sin θ

(1.17)

tan (90◦ − θ) = cot θ

(1.18)

Note that formulas (1.16)−(1.18) extend the Cofunction Theorem from Section 1.2 to all θ, not just acute angles. Similarly, formulas (1.10)−(1.12) and (1.13)−(1.15) give: sin (180◦ − θ) = sin θ

(1.19)

cos (180◦ − θ) = −cos θ

(1.20)

tan (180◦ − θ) = −tan θ

(1.21)

36

Chapter 1 • Right Triangle Trigonometry

§1.5

Notice that reflection around the y-axis is equivalent to reflection around the x-axis ( θ →

− θ) followed by a rotation of 180◦ (− θ → − θ + 180◦ = 180◦ − θ), as in Figure 1.5.7.

y

(− x, y)

( x, y)

180◦ − θ

r

r

θ

x

− θ

r

( x, − y)

Figure 1.5.7

Reflection of θ around the y-axis = 180◦ − θ

It may seem that these geometrical operations and formulas are not necessary for evalu-

ating the trigonometric functions, since we could just use a calculator. However, there are

two reasons for why they are useful. First, the formulas work for any angles, so they are

often used to prove general formulas in mathematics and other fields, as we will see later

in the text. Second, they can help in determining which angles have a given trigonometric

function value.

Example 1.27

Find all angles 0◦ ≤ θ < 360◦ such that sin θ = −0.682.

✄

Solution: Using the sin−1 button on a calculator with −0.682 as the input, we get θ = −43◦, which

✂

✁

is not between 0◦ and 360◦.7 Since θ = −43◦ is in QIV, its reflection 180◦ − θ around the y-axis will be in QIII and have the same sine value. But 180◦ − θ = 180◦ − (−43◦) = 223◦ (see Figure 1.5.8). Also, we

know that −43◦ and −43◦ +360◦ = 317◦ have the same trigonometric function values. So since angles

in QI and QII have positive sine values, we see that the only angles between 0◦ and 360◦ with a sine

of −0.682 are θ = 223◦ and 317◦ .

y

x

180◦ − θ = 223◦

θ = −43◦

r

r

(− x, y)

( x, y)

Figure 1.5.8

Reflection around the y-axis: −43◦ and 223◦

✄

7In Chapter 5 we will discuss why the sin−1 button returns that value.

✂

✁

Rotations and Reflections of Angles • Section 1.5

37

Exercises

1. Let θ = 32◦. Find the angle between 0◦ and 360◦ which is the

(a) reflection of θ around the x-axis

(b) reflection of θ around the y-axis

(c) reflection of θ around the origin

2. Repeat Exercise 1 with θ = 248◦.

3. Repeat Exercise 1 with θ = −248◦.

4. We proved formulas (1.4)-(1.6) for any angle θ in QI. Mimic that proof to show that the formulas hold for θ in QII.

5. Verify formulas (1.4)-(1.6) for θ on the coordinate axes, i.e. for θ = 0◦, 90◦, 180◦, 270◦.

6. In Example 1.26 we used the formulas involving θ + 90◦ to prove that the slopes of perpendicu-

lar lines are negative reciprocals. Show that this result can also be proved using the formulas

involving θ − 90◦. ( Hint: Only the last paragraph in that example needs to be modified. )

For Exercises 7 - 14, find all angles 0◦ ≤ θ < 360◦ which satisfy the given equation: 7. sin θ = 0.4226

8. sin θ = 0.1909

9. cos θ = 0.4226

10. sin θ = 0

11. tan θ = 0.7813

12. sin θ = −0.6294

13. cos θ = −0.9816

14. tan θ = −9.514

y

15. In our proof of the Pythagorean Theorem in Section 1.2, we

claimed that in a right triangle △ ABC it was possible to draw

B (0, b)

a line segment CD from the right angle vertex C to a point D

D

on the hypotenuse AB such that CD ⊥ AB. Use the picture on

the right to prove that claim. ( Hint: Notice how △ ABC is placed

on the x y-coordinate plane. What is the slope of the hypotenuse?

x

C (0, 0)

A ( a, 0)

What would be the slope of a line perpendicular to it? ) Also, find

the ( x, y) coordinates of the point D in terms of a and b.

16. It can be proved without using trigonometric functions that the slopes of perpendicular lines

are negative reciprocals. Let y = m 1 x + b 1 and y = m 2 x + b 2 be perpendicular lines (with nonzero slopes), as in the picture below. Use the picture to show that m 2 = − 1 .

m 1

( Hint: Think of similar triangles and the definition of slope. )

y

y = m 2 x + b 2

y = m 1 x + b 1

( x 2, y 2)

( x 3, y 2)

( x 2, y 1)

( x 3, y 1)

( x 1, y 1)

x

17. Prove formulas (1.19)-(1.21) by using formulas (1.10)−(1.12) and (1.13)−(1.15).

2 General Triangles

In Section 1.3 we saw how to solve a right triangle: given two sides, or one side and one

acute angle, we could find the remaining sides and angles. In each case we were actually

given three pieces of information, since we already knew one angle was 90◦.

For a general triangle, which may or may not have a right angle, we will again need three

pieces of information. The four cases are:

Case 1:

One side and two angles

Case 2:

Two sides and one opposite angle

Case 3:

Two sides and the angle between them

Case 4:

Three sides

Note that if we were given all three angles we could not determine the sides uniquely; by

similarity an infinite number of triangles have the same angles.

In this chapter we will learn how to solve a general triangle in all four of the above cases.

Though the methods described will work for right triangles, they are mostly used to solve

oblique triangles, that is, triangles which do not have a right angle. There are two types

of oblique triangles: an acute triangle has all acute angles, and an obtuse triangle has

one obtuse angle.

As we will see, Cases 1 and 2 can be solved using the law of sines, Case 3 can be solved

using either the law of cosines or the law of tangents, and Case 4 can be solved using the law

of cosines.

2.1 The Law of Sines

Theorem 2.1. Law of Sines: If a triangle has sides of lengths a, b, and c opposite the

angles A, B, and C, respectively, then

a

b

c

=

=

.

(2.1)

sin A

sin B

sin C

Note that by taking reciprocals, equation (2.1) can be written as

sin A

sin B

sin C

=

=

,

(2.2)

a

b

c

and it can also be written as a collection of three equations:

a

sin A

a

sin A

b

sin B

=

,

=

,

=

(2.3)

b

sin B

c

sin C

c

sin C

38

The Law of Sines • Section 2.1

39

Another way of stating the Law of Sines is: The sides of a triangle are proportional to the

sines of their opposite angles.

To prove the Law of Sines, let △ ABC be an oblique triangle. Then △ ABC can be acute, as

in Figure 2.1.1(a), or it can be obtuse, as in Figure 2.1.1(b). In each case, draw the altitude 1

from the vertex at C to the side AB. In Figure 2.1.1(a) the altitude lies inside the triangle,

while in Figure 2.1.1(b) the altitude lies outside the triangle.

C

C

b

a

b

h

h

a

180◦ − B

c

A

B

c

A

B

(a) Acute triangle

(b) Obtuse triangle

Figure 2.1.1

Proof of the Law of Sines for an oblique triangle △ ABC

Let h be the height of the altitude. For each triangle in Figure 2.1.1, we see that

h = sin A

(2.4)

b

and

h = sin B

(2.5)

a

(in Figure 2.1.1(b), ha = sin (180◦ − B) = sin B by formula (1.19) in Section 1.5). Thus, solving for h in equation (2.5) and substituting that into equation (2.4) gives

a sin B = sin A ,

(2.6)

b

and so putting a and A on the left side and b and B on the right side, we get

a

b

=

.

(2.7)

sin A

sin B

By a similar argument, drawing the altitude from A to BC gives

b

c

=

,

(2.8)

sin B

sin C

so putting the last two equations together proves the theorem.

QED

Note that we did not prove the Law of Sines for right triangles, since it turns out (see

Exercise 12) to be trivially true for that case.

1Recall from geometry that an altitude of a triangle is a perpendicular line segment from any vertex to the line

containing the side opposite the vertex.

40

Chapter 2 • General Triangles

§2.1

Example 2.1

Case 1: One side and two angles.

C = 75◦

Solve the triangle △ ABC given a = 10, A = 41◦, and C = 75◦.

b

a = 10

Solution: We can find the third angle by subtracting the other two angles

from 180◦, then use the law of sines to find the two unknown sides. In this

example we need to find B, b, and c. First, we see that

c

A = 41◦

B

B = 180◦ − A − C = 180◦ − 41◦ − 75◦

⇒

B = 64◦ .

So by the Law of Sines we have

b

a

a sin B

10 sin 64◦

=

⇒

b =

=

⇒

b = 13.7 , and

sin B

sin A

sin A

sin 41◦

c

a

a sin C

10 sin 75◦

=

⇒

c =

=

⇒

c = 14.7 .

sin C

sin A

sin A

sin 41◦

Example 2.2

Case 2: Two sides and one opposite angle.

C

Solve the triangle △ ABC given a = 18, A = 25◦, and b = 30.

b = 30

a = 18

Solution: In this example we know the side a and its opposite angle A,

and we know the side b. We can use the Law of Sines to find the other

c

A = 25◦

B

opposite angle B, then find the third angle C by subtracting A and B from

180◦, then use the law of sines to find the third side c. By the Law of Sines, we have

sin B

sin A

b sin A

30 sin 25◦

=

⇒

sin B =

=

⇒

sin B = 0.7044 .

b

a

a

18

✄

Using the sin−1 button on a calculator gives B = 44.8◦. However, recall from Section 1.5 that

✂

✁

sin (180◦ − B) = sin B. So there is a second possible solution for B, namely 180◦ − 44.8◦ = 135.2◦.

Thus, we have to solve twice for C and c : once for B = 44.8◦ and once for B = 135.2◦:

B = 44.8◦

B = 135.2◦

C = 180◦ − A − B = 180◦ − 25◦ − 44.8◦ = 110.2◦

C = 180◦ − A − B = 180◦ − 25◦ − 135.2◦ = 19.8◦

c

a

a sin C

18 sin 110.2◦

c

a

a sin C

18 sin 19.8◦

=

⇒ c =

=

=

⇒ c =

=

sin C

sin A

sin A

sin 25◦

sin C

sin A

sin A

sin 25◦

⇒ c = 40

⇒ c = 14.4

Hence, B = 44.8◦, C = 110.2◦, c = 40 and B = 135.2◦, C = 19.8◦, c = 14.4 are the two possible sets of solutions. This means that there are two possible triangles, as shown in Figure 2.1.2.

C = 110.2◦

C = 19.8◦

b = 30

a = 18

b = 30

a = 18

A = 25◦

B = 44.8◦

A = 25◦

B = 135.2◦

c = 40

c = 14.4

(a) B = 44.8◦

(b) B = 135.2◦

Figure 2.1.2

Two possible solutions

The Law of Sines • Section 2.1

41

In Example 2.2 we saw what is known as the ambiguous case. That is, there may be more

than one solution. It is also possible for there to be exactly one solution or no solution at all.

Example 2.3

Case 2: Two sides and one opposite angle.

Solve the triangle △ ABC given a = 5, A = 30◦, and b = 12.

Solution: By the Law of Sines, we have

sin B

sin A

b sin A

12 sin 30◦

=

⇒

sin B =

=

⇒

sin B = 1.2 ,

b

a

a

5

which is impossible since |sin B| ≤ 1 for any angle B. Thus, there is no solution .

There is a way to determine how many solutions a triangle has in Case 2. For a triangle

△ ABC, suppose that we know the sides a and b and the angle A. Draw the angle A and

the side b, and imagine that the side a is attached at the vertex at C so that it can “swing”

freely, as indicated by the dashed arc in Figure 2.1.3 below.

C

C

C

a

b

h

b

b

a

h a

h

a

A

B

c

A

B

A

B

B

(a) a < h: No solution

(b) a = h: One solution

(c) h < a < b: Two solutions

C

b

a

b

c

A

B

(d) a ≥ b: One solution

Figure 2.1.3

The ambiguous case when A is acute

If A is acute, then the altitude from C to AB has height h = b sin A. As we can see in

Figure 2.1.3(a)-(c), there is no solution when a < h (this was the case in Example 2.3); there is exactly one solution - namely, a right triangle - when a = h; and there are two solutions

when h < a < b (as was the case in Example 2.2). When a ≥ b there is only one solution, even though it appears from Figure 2.1.3(d) that there may be two solutions, since the dashed arc

intersects the horizontal line at two points. However, the point of intersection to the left of A

in Figure 2.1.3(d) can not be used to determine B, since that would make A an obtuse angle,

and we assumed that A was acute.

If A is not acute (i.e. A is obtuse or a right angle), then the situation is simpler: there is

no solution if a ≤ b, and there is exactly one solution if a > b (see Figure 2.1.4).

42

Chapter 2 • General Triangles

§2.1

C

C

a

a

b

b

A

B

A

B

(a) a ≤ b: No solution

(b) a > b: One solution

Figure 2.1.4

The ambiguous case when A ≥ 90◦

Table 2.1 summarizes the ambiguous case of solving △ ABC when given a, A, and b. Of course, the letters can be interchanged, e.g. replace a and A by c and C, etc.

Table 2.1

Summary of the ambiguous case

0◦ < A < 90◦

90◦ ≤ A < 180◦

a < b sin A :

No solution

a ≤ b :

No solution

a = b sin A :

One solution

a > b :

One solution

b sin A < a < b :

Two solutions

a ≥ b :

One solution

There is an interesting geometric consequence of the Law of Sines. Recall from Section 1.1

that in a right triangle the hypotenuse is the largest side. Since a right angle is the largest

angle in a right triangle, this means that the largest side is opposite the largest angle. What

the Law of Sines does is generalize this to any triangle:

In any triangle, the largest side is opposite the largest angle.

To prove this, let C be the largest angle in a triangle △ ABC. If C = 90◦ then we already

know that its opposite side c is the largest side. So we just need to prove the result for when

C is acute and for when C is obtuse. In both cases, we have A ≤ C and B ≤ C. We will first show that sin A ≤ sin C and sin B ≤ sin C.

y

If C is acute, then A and B are also acute. Since A ≤ C, imagine

( x 2, y 2)

that A is in standard position in the x y-coordinate plane and that

we rotate the terminal side of A counterclockwise to the terminal

r

side of the larger angle C, as in Figure 2.1.5. If we pick points

C

( x

( x 1, y 1)

1, y 1) and ( x 2, y 2) on the terminal sides of A and C, respectively,

r

so that their distance to the origin is the same number r, then we

A

x

see from the picture that y 1 ≤ y 2, and hence

Figure 2.1.5

y 1

y 2

sin A =

≤

= sin C .

r

r

By a similar argument, B ≤ C implies that sin B ≤ sin C. Thus, sin A ≤ sin C and sin B ≤

sin C when C is acute. We will now show that these inequalities hold when C is obtuse.

The Law of Sines • Section 2.1

43

If C is obtuse, then 180◦ − C is acute, as are A and B. If A > 180◦ − C then A + C > 180◦, which is impossible. Thus, we must have A ≤ 180◦− C. Likewise, B ≤ 180◦− C. So by what we

showed above for acute angles, we know that sin A ≤ sin (180◦− C) and sin B ≤ sin (180◦− C).

But we know from Section 1.5 that sin C = sin (180◦ − C). Hence, sin A ≤ sin C and sin B ≤

sin C when C is obtuse.

Thus, sin A ≤ sin C if C is acute or obtuse, so by the Law of Sines we have

a

sin A

sin C

a

=

≤

= 1

⇒

≤ 1

⇒

a ≤ c .

c

sin C

sin C

c

By a similar argument, b ≤ c. Thus, a ≤ c and b ≤ c, i.e. c is the largest side.

QED

Exercises

For Exercises 1-9, solve the triangle △ ABC.

1. a = 10, A = 35◦, B = 25◦

2. b = 40, B = 75◦, c = 35

3. A = 40◦, B = 45◦, c = 15

4. a = 5, A = 42◦, b = 7

5. a = 40, A = 25◦, c = 30

6. a = 5, A = 47◦, b = 9

7. a = 12, A = 94◦, b = 15

8. a = 15, A = 94◦, b = 12

9. a = 22, A = 50◦, c = 27

10. Draw a circle with a radius of 2 inches and inscribe a triangle inside the circle. Use a ruler and

a protractor to measure the sides a, b, c and the angles A, B, C of the triangle. The Law of Sines says that the ratios

a

,

b

,

c

are equal. Verify this for your triangle. What relation does

sin A

sin B

sin C

that common ratio have to the diameter of your circle?

11. An observer on the ground measures an angle of inclination of 30◦ to an approaching airplane,

and 10 seconds later measures an angle of inclination of 55◦. If the airplane is flying at a constant

speed and at a steady altitude of 6000 ft in a straight line directly over the observer, find the speed

of the airplane in miles per hour. (Note: 1 mile = 5280 ft)

10 seconds pass

✈

✈

55◦

6000 ft

30◦

12. Prove the Law of Sines for right triangles. ( Hint: One of the angles is known. )

a ± b

sin A ± sin B

13. For a triangle △ ABC, show that

=

.

c

sin C

a

sin ( B + C)

14. For a triangle △ ABC, show that

=

.

c

sin C

15. One diagonal of a parallelogram is 17 cm long and makes angles of 36◦ and 15◦ with the sides.

Find the lengths of the sides.

16. Explain why in Case 1 (one side and two angles) there is always exactly one solution.

44

Chapter 2 • General Triangles

§2.2

2.2 The Law of Cosines

We will now discuss how to solve a triangle in Case 3: two sides and the angle between them.

First, let us see what happens when we try to use the Law of Sines for this case.

Example 2.4

Case 3: Two sides and the angle between them.

C

Solve the triangle △ ABC given A = 30◦, b = 4, and c = 5.

b = 4

a

Solution: Using the Law of Sines, we have

A

c

= 30◦

= 5

B

a

4

5

=

=

,

sin 30◦

sin B

sin C

where each of the equations has two unknown parts, making the problem impossible to solve. For

example, to solve for a we could use the equation

4

to solve for sin B in terms of sin C and

sin B =

5

sin C

substitute that into the equation

a

. But that would just result in the equation

a

sin 30◦ =

4

sin B

sin 30◦ =

5

, which we already knew and which still has two unknowns!

sin C

Thus, this problem can not be solved using the Law of Sines.

To solve the triangle in the above example, we can use the Law of Cosines:

Theorem 2.2. Law of Cosines: If a triangle has sides of lengths a, b, and c opposite the

angles A, B, and C, respectively, then

a 2 = b 2 + c 2 − 2 bc cos A ,

(2.9)

b 2 = c 2 + a 2 − 2 ca cos B ,

(2.10)

c 2 = a 2 + b 2 − 2 ab cos C .

(2.11)

To prove the Law of Cosines, let △ ABC be an oblique triangle. Then △ ABC can be acute,

as in Figure 2.2.1(a), or it can be obtuse, as in Figure 2.2.1(b). In each case, draw the altitude from the vertex at C to the side AB. In Figure 2.2.1(a) the altitude divides AB into two line segments with lengths x and c − x, while in Figure 2.2.1(b) the altitude extends the side AB

by a distance x. Let h be the height of the altitude.

C

C

b

a

b

h

h

a

c − x

x

180◦ − B

c

A

B

c

A

B

x

(a) Acute triangle

(b) Obtuse triangle

Figure 2.2.1

Proof of the Law of Cosines for an oblique triangle △ ABC

The Law of Cosines • Section 2.2

45

For each triangle in Figure 2.2.1, we see by the Pythagorean Theorem that

h 2 = a 2 − x 2

(2.12)

and likewise for the acute triangle in Figure 2.2.1(a) we see that

b 2 = h 2 + ( c − x)2 .

(2.13)

Thus, substituting the expression for h 2 in equation (2.12) into equation (2.13) gives b 2 = a 2 − x 2 + ( c − x)2

= a 2 − x 2 + c 2 − 2 cx + x 2

= a 2 + c 2 − 2 cx .

But we see from Figure 2.2.1(a) that x = a cos B, so

b 2 = a 2 + c 2 − 2 ca cos B .

(2.14)

And for the obtuse triangle in Figure 2.2.1(b) we see that

b 2 = h 2 + ( c + x)2 .

(2.15)

Thus, substituting the expression for h 2 in equation (2.12) into equation (2.15) gives b 2 = a 2 − x 2 + ( c + x)2

= a 2 − x 2 + c 2 + 2 cx + x 2

= a 2 + c 2 + 2 cx .

But we see from Figure 2.2.1(a) that x = a cos (180◦ − B), and we know from Section 1.5 that cos (180◦ − B) = −cos B. Thus, x = − a cos B and so

b 2 = a 2 + c 2 − 2 ca cos B .

(2.16)

So for both acute and obtuse triangles we have proved formula (2.10) in the Law of Cosines.

Notice that the proof was for B acute and obtuse. By similar arguments for A and C we get

the other two formulas.

QED

Note that we did not prove the Law of Cosines for right triangles, since it turns out (see

Exercise 15) that all three formulas reduce to the Pythagorean Theorem for that case. The

Law of Cosines can be viewed as a generalization of the Pythagorean Theorem.

Also, notice that it suffices to remember just one of the three formulas (2.9)-(2.11), since the other two can be obtained by “cycling” through the letters a, b, and c. That is, replace a

by b, replace b by c, and replace c by a (likewise for the capital letters). One cycle will give

you the second formula, and another cycle will give you the third.

46

Chapter 2 • General Triangles

§2.2

The angle between two sides of a triangle is often called the included angle. Notice in

the Law of Cosines that if two sides and their included angle are known (e.g. b, c, and A),

then we have a formula for the square of the third side.

We will now solve the triangle from Example 2.4.

Example 2.5

Case 3: Two sides and the angle between them.

C

Solve the triangle △ ABC given A = 30◦, b = 4, and c = 5.

b = 4

a

Solution: We will use the Law of Cosines to find a, use it again to find

B, then use C = 180◦ − A − B. First, we have

A

c

= 30◦

= 5

B

a 2 = b 2 + c 2 − 2 bc cos A

= 42 + 52 − 2(4)(5) cos 30◦ = 6.36

⇒

a = 2.52 .

Now we use the formula for b 2 to find B:

c 2 + a 2 − b 2

b 2 = c 2 + a 2 − 2 ca cos B

⇒

cos B =

2 ca

52 + (2.52)2 − 42

⇒

cos B =

= 0.6091

2(5)(2.52)

⇒

B = 52.5◦

Thus, C = 180◦ − A − B = 180◦ − 30◦ − 52.5◦ ⇒ C = 97.5◦ .

Notice in Example 2.5 that there was only one solution. For Case 3 this will always be

true: when given two sides and their included angle, the triangle will have exactly one

solution. The reason is simple: when joining two line segments at a common vertex to form

an angle, there is exactly one way to connect their free endpoints with a third line segment,

regardless of the size of the angle.

You may be wondering why we used the Law of Cosines a second time in Example 2.5, to

find the angle B. Why not use the Law of Sines, which has a simpler formula? The reason

is that using the cosine function eliminates any ambiguity: if the cosine is positive then the

angle is acute, and if the cosine is negative then the angle is obtuse. This is in contrast

to using the sine function; as we saw in Section 2.1, both an acute angle and its obtuse

supplement have the same positive sine.

To see this, suppose that we had used the Law of Sines to find B in Example 2.5:

b sin A

4 sin 30◦

sin B =

=

= 0.7937

⇒

B = 52.5◦ or 127.5◦

a

2.52

How would we know which answer is correct? We could not immediately rule out B = 127.5◦

as too large, since it would make A + B = 157.5◦ < 180◦ and so C = 22.5◦, which seems like

it could be a valid solution. However, this solution is impossible. Why? Because the largest

side in the triangle is c = 5, which (as we learned in Section 2.1) means that C has to be the

largest angle. But C = 22.5◦ would not be the largest angle in this solution, and hence we

have a contradiction.

The Law of Cosines • Section 2.2

47

It remains to solve a triangle in Case 4, i.e. given three sides. We will now see how to use

the Law of Cosines for that case.

Example 2.6

Case 4: Three sides.

C

Solve the triangle △ ABC given a = 2, b = 3, and c = 4.

b = 3

a = 2

Solution: We will use the Law of Cosines to find B and C, then use A =

180◦ − B − C. First, we use the formula for b 2 to find B:

A

c = 4

B

c 2 + a 2 − b 2

b 2 = c 2 + a 2 − 2 ca cos B

⇒

cos B =

2 ca

42 + 22 − 32

⇒

cos B =

= 0.6875

2(4)(2)

⇒

B = 46.6◦

Now we use the formula for c 2 to find C:

a 2 + b 2 − c 2

c 2 = a 2 + b 2 − 2 ab cos C

⇒

cos C =

2 ab

22 + 32 − 42

⇒

cos C =

= −0.25

2(2)(3)

⇒

C = 104.5◦

Thus, A = 180◦ − B − C = 180◦ − 46.6◦ − 104.5◦ ⇒ A = 28.9◦ .

It may seem that there is always a solution in Case 4 (given all three sides), but that is

not true, as the following example shows.

Example 2.7

Case 4: Three sides.

C

Solve the triangle △ ABC given a = 2, b = 3, and c = 6.

b = 3

a = 2

Solution: If we blindly try to use the Law of Cosines to find A, we get

A

B

c

b 2

32

= 6

+ c 2 − a 2

+ 62 − 22

a 2 = b 2 + c 2 − 2 bc cos A

⇒

cos A =

=

= 1.139 ,

2 bc

2(3)(6)

which is impossible since |cos A| ≤ 1. Thus, there is no solution .

We could have saved ourselves some effort by recognizing that the length

b = 3

a = 2

of one of the sides ( c = 6) is greater than the sums of the lengths of the

remaining sides ( a = 2 and b = 3), which (as the picture on the right shows)

c = 6

is impossible in a triangle.

The Law of Cosines can also be used to solve triangles in Case 2 (two sides and one op-

posite angle), though it is less commonly used for that purpose than the Law of Sines. The

following example gives an idea of how to do this.

48

Chapter 2 • General Triangles

§2.2

Example 2.8

Case 2: Two sides and one opposite angle.

Solve the triangle △ ABC given a = 18, A = 25◦, and b = 30.

Solution: In Example 2.2 from Section 2.1 we used the Law of Sines to show that there are two sets

of solutions for this triangle: B = 44.8◦, C = 110.2◦, c = 40 and B = 135.2◦, C = 19.8◦, c = 14.4. To solve this using the Law of Cosines, first find c by using the formula for a 2:

a 2 = b 2 + c 2 − 2 bc cos A

⇒

182 = 302 + c 2 − 2(30) c cos 25◦

⇒

c 2 − 54.38 c + 576 = 0 ,

which is a quadratic equation in c, so we know that it can have either zero, one, or two real roots

(corresponding to the number of solutions in Case 2). By the quadratic formula, we have

54.38 ±

(54.38)2 − 4(1)(576)

c =

= 40 or 14.4 .

2(1)

Note that these are the same values for c that we found before. For c = 40 we get

c 2 + a 2 − b 2

402 + 182 − 302

cos B =

=

= 0.7111

⇒

B = 44.7◦

⇒

C = 110.3◦ ,

2 ca

2(40)(18)

which is close to what we found before (the small difference being due to different rounding). The

other solution set can be obtained similarly.

Like the Law of Sines, the Law of Cosines can be used to prove some geometric facts, as

in the following example.

Example 2.9

Use the Law of Cosines to prove that the sum of the squares of the

D

b

diagonals of any parallelogram equals the sum of the squares of the

sides.

c

a

a

Solution: Let a and b be the lengths of the sides, and let the diago-

nals opposite the angles C and D have lengths c and d, respectively,

d

as in Figure 2.2.2. Then we need to show that

C

b

c 2 + d 2 = a 2 + b 2 + a 2 + b 2 = 2( a 2 + b 2) .

Figure 2.2.2

By the Law of Cosines, we know that

c 2 = a 2 + b 2 − 2 ab cos C , and

d 2 = a 2 + b 2 − 2 ab cos D .

By properties of parallelograms, we know that D = 180◦ − C, so

d 2 = a 2 + b 2 − 2 ab cos (180◦ − C)

= a 2 + b 2 + 2 ab cos C ,

since cos (180◦ − C) = −cos C. Thus,

c 2 + d 2 = a 2 + b 2 − 2 ab cos C + a 2 + b 2 + 2 ab cos C

= 2( a 2 + b 2) .

QED

The Law of Cosines • Section 2.2

49

Exercises

For Exercises 1-6, solve the triangle △ ABC.

1. A = 60◦, b = 8, c = 12

2. A = 30◦, b = 4, c = 6

3. a = 7, B = 60◦, c = 9

4. a = 7, b = 3, c = 9

5. a = 6, b = 4, c = 1

6. a = 11, b = 13, c = 16

7. The diagonals of a parallelogram intersect at a 42◦ angle and have lengths of 12 and 7 cm. Find

the lengths of the sides of the parallelogram. ( Hint: The diagonals bisect each other. )

8. Two trains leave the same train station at the same time, moving along straight tracks that form

a 35◦ angle. If one train travels at an average speed of 100 mi/hr and the other at an average

speed of 90 mi/hr, how far apart are the trains after half an hour?

9. Three circles with radii of 4, 5, and 6 cm, respectively, are tangent to each other externally. Find

the angles of the triangle whose vertexes are the centers of the circles.

10. Find the length x of the diagonal of the quadrilateral in Figure 2.2.3 below.

4

3.5

60◦

x

5

3

6

2

5.5

Figure 2.2.3

Exercise 10

Figure 2.2.4

Exercise 11

11. Two circles of radii 5 and 3 cm, respectively, intersect at two points. At either point of intersec-

tion, the tangent lines to the circles form a 60◦ angle, as in Figure 2.2.4 above. Find the distance

between the centers of the circles.

12. Use the Law of Cosines to show that for any triangle △ ABC, c 2 < a 2+ b 2 if C is acute, c 2 > a 2+ b 2

if C is obtuse, and c 2 = a 2 + b 2 if C is a right angle.

13. Show that for any triangle △ ABC,

cos A

cos B

cos C

a 2 + b 2 + c 2

+

+

=

.

a

b

c

2 abc

14. Show that for any triangle △ ABC,

a 2 ( b + c − a) + b 2 ( a + c − b) + c 2 ( a + b − c) cos A + cos B + cos C =

.

2 abc

What do the terms in parentheses represent geometrically? Use your answer to explain why

cos A + cos B + cos C > 0 for any triangle, even if one of the cosines is negative.2

15. Prove the Law of Cosines (i.e. formulas (2.9)-(2.11)) for right triangles.

16. Recall from elementary geometry that a median of a triangle is a line segment from any vertex

to the midpoint of the opposite side. Show that the sum of the squares of the three medians of a

triangle is 3⁄4 the sum of the squares of the sides.

2It turns out that 1 < cos A + cos B + cos C ≤ 3/2 for any triangle, as we will see later.

50

Chapter 2 • General Triangles

§2.2

17. The Dutch astronomer and mathematician Willebrord Snell (1580-1626) wrote the Law of Cosines

as

2 ab

1

=

c 2 − ( a − b)2

1 − cos C

in his trigonometry text Doctrina triangulorum (published a year after his death). Show that this

formula is equivalent to formula (2.11) in our statement of the Law of Cosines.

18. Suppose that a satellite in space, an earth station, and the center of the earth all lie in the same

plane. Let re be the radius of the earth, let rs be the distance from the center of the earth to the

satellite (called the orbital radius of the satellite), and let d be the distance from the earth station

to the satellite. Let E be the angle of elevation from the earth station to the satellite, and let γ

and ψ be the angles shown in Figure 2.2.5.

satellite

✸

local horizontal

d

rs

E

ψ

γ

❚

r

earth station

e

center

of earth

Figure 2.2.5

Use the Law of Cosines to show that

r

2

e

r e

d = rs

1 +

− 2

cos γ ,

rs

rs

and then use E = ψ − 90◦ and the Law of Sines to show that

sin γ

cos E =

.

r

2

e

r e

1 +

− 2

cos γ

rs

rs

Note: This formula allows the angle of elevation E to be calculated from the coordinates of the

earth station and the subsatellite point (where the line from the satellite to the center of the earth

crosses the surface of the earth).3

3See pp. 22-25 in T. PRATT AND C.W. BOSTIAN, Satellite Communications, New York: John Wiley & Sons,

1986.

The Law of Tangents • Section 2.3

51

2.3 The Law of Tangents

We have shown how to solve a triangle in all four cases discussed at the beginning of this

chapter. An alternative to the Law of Cosines for Case 3 (two sides and the included angle)

is the Law of Tangents:

Theorem 2.3. Law of Tangents: If a triangle has sides of lengths a, b, and c opposite the

angles A, B, and C, respectively, then

a − b

tan 1 ( A − B)

=

2

,

(2.17)

a + b

tan 1 ( A

2

+ B)

b − c

tan 1 ( B − C)

=

2

,

(2.18)

b + c

tan 1 ( B

2

+ C)

c − a

tan 1 ( C − A)

=

2

.

(2.19)

c + a

tan 1 ( C

2

+ A)

Note that since tan (− θ) = −tan θ for any angle θ, we can switch the order of the letters in

each of the above formulas. For example, we can rewrite formula (2.17) as

b − a

tan 1 ( B − A)

=

2

,

(2.20)

b + a

tan 1 ( B

2

+ A)

and similarly for the other formulas. If a > b, then it is usually more convenient to use

formula (2.17), while formula (2.20) is more convenient when b > a.

Example 2.10

Case 3: Two sides and the included angle.

C = 96◦

Solve the triangle △ ABC given a = 5, b = 3, and C = 96◦.

b = 3

a = 5

Solution: A + B + C = 180◦, so A + B = 180◦ − C = 180◦ − 96◦ = 84◦. Thus, by

the Law of Tangents,

c

A

B

a − b

tan 1 ( A − B)

5 − 3

tan 1 ( A − B)

=

2

⇒

=

2

a + b

tan 1 ( A

5 + 3

tan 1 (84◦)

2

+ B)

2

⇒

tan 1 ( A

tan 42◦

2

− B) = 28

= 0.2251

⇒

1 ( A

2

− B) = 12.7◦

⇒

A − B = 25.4◦ .

We now have two equations involving A and B, which we can solve by adding the equations:

A − B = 25.4◦

A + B = 84◦

−− − − − − − −

2 A

= 109.4◦

⇒

A = 54.7◦

⇒

B = 84◦ − 54.7◦

⇒

B = 29.3◦

We can find the remaining side c by using the Law of Sines:

a sin C

5 sin 96◦

c =

=

⇒

c = 6.09

sin A

sin 54.7◦

52

Chapter 2 • General Triangles

§2.3

Note that in any triangle △ ABC, if a = b then A = B (why?), and so both sides of formula

(2.17) would be 0 (since tan 0◦ = 0). This means that the Law of Tangents is of no help in

Case 3 when the two known sides are equal. For this reason, and perhaps also because of the

somewhat unusual way in which it is used, the Law of Tangents seems to have fallen out of

favor in trigonometry books lately. It does not seem to have any advantages over the Law

of Cosines, which works even when the sides are equal, requires slightly fewer steps, and is

perhaps more straightforward.4

Related to the Law of Tangents are Mollweide’s equations:5

Mollweide’s equations: For any triangle △ ABC,

a − b

sin 1 ( A − B)

=

2

, and

(2.21)

c

cos 1 C

2

a + b

cos 1 ( A − B)

=

2

.

(2.22)

c

sin 1 C

2

Note that all six parts of a triangle appear in both of Mollweide’s equations. For this

reason, either equation can be used to check a solution of a triangle. If both sides of the

equation agree (more or less), then we know that the solution is correct.

Example 2.11

Use one of Mollweide’s equations to check the solution of the triangle from Example 2.10.

Solution: Recall that the full solution was a = 5, b = 3, c = 6.09, A = 54.7◦, B = 29.3◦, and C = 96◦.

We will check this with equation (2.21):

a − b

sin 1 ( A − B)

=

2

c

cos 1 C

2

5 − 3

sin 1 (54.7◦ − 29.3◦)

=

2

6.09

cos 1 (96◦)

2

2

sin 12.7◦

=

6.09

cos 48◦

0.3284 = 0.3285 ✓

The small difference (≈ 0.0001) is due to rounding errors from the original solution, so we can con-

clude that both sides of the equation agree, and hence the solution is correct.

4Before the advent of electronic calculators, the Law of Tangents was more popular than it is today since

it lent itself better than the Law of Cosines to what was known as logarithmic computation. In those days,

computations with large numbers were handled by taking logarithms and looking up values in a logarithm table.

Ratios (such as in the Law of Tangents and the Law of Sines) could be replaced by differences of logarithms,

making computation easier.

5Named after the German astronomer and mathematician Karl Mollweide (1774-1825).

The Law of Tangents • Section 2.3

53

Example 2.12

Can a triangle have the parts a = 6, b = 7, c = 9, A = 55◦, B = 60◦, and C = 65◦ ?

Solution: Before using Mollweide’s equations, simpler checks are that the angles add up to 180◦ and

that the smallest and largest sides are opposite the smallest and largest angles, respectively. In this

case all those conditions hold. So check with Mollweide’s equation (2.22):

a + b

cos 1 ( A − B)

=

2

c

sin 1 C

2

6 + 7

cos 1 (55◦ − 60◦)

=

2

9

sin 1 (65◦)

2

13

cos (−2.5◦)

=

9

sin 32.5◦

1.44 = 1.86 ✗

Here the difference is far too large, so we conclude that there is no triangle with these parts.

We will prove the Law of Tangents and Mollweide’s equations in Chapter 3, where we will

be able to supply brief analytic proofs.6

Exercises

For Exercises 1-3, use the Law of Tangents to solve the triangle △ ABC.

1. a = 12, b = 8, C = 60◦

2. A = 30◦, b = 4, c = 6

3. a = 7, B = 60◦, c = 9

For Exercises 4-6, check if it is possible for a triangle to have the given parts.

4. a = 5, b = 7, c = 10, A = 27.7◦, B = 40.5◦, C = 111.8◦

5. a = 3, b = 7, c = 9, A = 19.2◦, B = 68.2◦, C = 92.6◦

6. a = 6, b = 9, c = 9, A = 39◦, B = 70.5◦, C = 70.5◦

7. Let △ ABC be a right triangle with C = 90◦. Show that tan 1 ( A

.

2

− B) = a− b

a+ b

8. For any triangle △ ABC, show that tan 1 ( A

cot 1 C .

2

− B) = a− b

a+ b

2

a sin B

9. For any triangle △ ABC, show that tan A =

. ( Hint: Draw the altitude from the vertex

c − a cos B

C to AB. ) Notice that this formula provides another way of solving a triangle in Case 3 (two sides

and the included angle).

10. For any triangle △ ABC, show that c = b cos A + a cos B . This is another check of a triangle.

11. If b cos A = a cos B , show that the triangle △ ABC is isosceles.

12. Let ABCD be a quadrilateral which completely contains its two diagonals. The quadrilateral

has eight parts: four sides and four angles. What is the smallest number of parts that you would

need to know to solve the quadrilateral? Explain your answer.

6There are (complex) geometric proofs of the Law of Tangents and Mollweide’s equations. See pp. 96-98 in P.R.

RIDER, Plane and Spherical Trigonometry, New York: The Macmillan Company, 1942.

54

Chapter 2 • General Triangles

§2.4

2.4 The Area of a Triangle

In elementary geometry you learned that the area of a triangle is one-half the base times the

height. We will now use that, combined with some trigonometry, to derive more formulas for

the area when given various parts of the triangle.

Case 1: Two sides and the included angle.

Suppose that we have a triangle △ ABC, in which A can be either acute, a right angle, or

obtuse, as in Figure 2.4.1. Assume that A, b, and c are known.

C

C

C

b

h

a

a

a

b

h

h

b

A

c

B

A

c

B

A

c

B

(a) A acute

(b) A = 90◦

(c) A obtuse

Figure 2.4.1

Area of △ ABC

In each case we draw an altitude of height h from the vertex at C to AB, so that the area

(which we will denote by the letter K ) is given by K = 1 hc. But we see that h

2

= b sin A

in each of the triangles (since h = b and sin A = sin 90◦ = 1 in Figure 2.4.1(b), and h =

b sin (180◦ − A) = b sin A in Figure 2.4.1(c)). We thus get the following formula: Area = K = 1 bc sin A

(2.23)

2

The above formula for the area of △ ABC is in terms of the known parts A, b, and c. Similar

arguments for the angles B and C give us:

Area = K = 1 ac sin B

(2.24)

2

Area = K = 1 ab sin C

(2.25)

2

Notice that the height h does not appear explicitly in these formulas, although it is implicitly

there. These formulas have the advantage of being in terms of parts of the triangle, without

having to find h separately.

Example 2.13

Find the area of the triangle △ ABC given

C

A = 33◦, b = 5, and c = 7.

b = 5

a

Solution: Using formula (2.23), the area K is given by:

A

c

= 33◦

= 7

B

K = 1 bc sin A

2

= 1 (5)(7) sin 33◦

2

K = 9.53

The Area of a Triangle • Section 2.4

55

Case 2: Three angles and any side.

Suppose that we have a triangle △ ABC in which one side, say, a, and all three angles are

known.7 By the Law of Sines we know that

a sin C

c =

,

sin A

so substituting this into formula (2.24) we get:

a 2 sin B sin C

Area = K =

(2.26)

2 sin A

Similar arguments for the sides b and c give us:

b 2 sin A sin C

Area = K =

(2.27)

2 sin B

c 2 sin A sin B

Area = K =

(2.28)

2 sin C

Example 2.14

Find the area of the triangle △ ABC given

C = 40◦

A = 115◦, B = 25◦, C = 40◦, and a = 12.

a = 12

Solution: Using formula (2.26), the area K is given by:

b

a 2

A

sin B sin C

= 115◦ c

B = 25◦

K =

2 sin A

122 sin 25◦ sin 40◦

=

2 sin 115◦

K = 21.58

Case 3: Three sides.

Suppose that we have a triangle △ ABC in which all three sides are known. Then Heron’s

formula 8 gives us the area:

Heron’s formula: For a triangle △ ABC with sides a, b, and c, let s = 1 ( a

2

+ b + c) (i.e.

2 s = a + b + c is the perimeter of the triangle). Then the area K of the triangle is

Area = K =

s ( s − a)( s − b)( s − c) .

(2.29)

To prove this, first remember that the area K is one-half the base times the height. Using

c as the base and the altitude h as the height, as before in Figure 2.4.1, we have K = 1 hc.

2

Squaring both sides gives us

K 2 = 1 h 2 c 2 .

(2.30)

4

7Note that this is equivalent to knowing just two angles and a side (why?).

8Due to the ancient Greek engineer and mathematician Heron of Alexandria (c. 10-70 A.D.).

56

Chapter 2 • General Triangles

§2.4

In Figure 2.4.2, let D be the point where the altitude touches AB (or its extension).

C

C

b

h

a

a

h

b

A

c D

B

A

c

D

B

(a) A acute

(b) A obtuse

Figure 2.4.2

Proof of Heron’s formula

By the Pythagorean Theorem, we see that h 2 = b 2 − ( AD)2. In Figure 2.4.2(a), we see that AD = b cos A. And in Figure 2.4.2(b) we see that AD = b cos (180◦ − A) = − b cos A. Hence, in either case we have ( AD)2 = b 2 (cos A)2, and so

h 2 = b 2 − b 2 (cos A)2 = b 2 (1 − (cos A)2) = b 2 (1 + cos A)(1 − cos A) .

(2.31)

(Note that the above equation also holds when A = 90◦ since cos 90◦ = 0 and h = b). Thus,

substituting equation (2.31) into equation (2.30), we have

K 2 = 1 b 2 c 2 (1

4

+ cos A)(1 − cos A) .

(2.32)

By the Law of Cosines we know that

b 2 + c 2 − a 2

2 bc + b 2 + c 2 − a 2

( b + c)2 − a 2

(( b + c) + a)(( b + c) − a)

1 + cos A = 1 +

=

=

=

2 bc

2 bc

2 bc

2 bc

( a + b + c)( b + c − a)

=

,

2 bc

and similarly

b 2 + c 2 − a 2

2 bc − b 2 − c 2 + a 2

a 2 − ( b − c)2

( a − ( b − c))( a + ( b − c))

1 − cos A = 1 −

=

=

=

2 bc

2 bc

2 bc

2 bc

( a − b + c)( a + b − c)

=

.

2 bc

Thus, substituting these expressions into equation (2.32), we have

( a − b + c)( a + b − c)

K 2 = 1 b 2 c 2 ( a + b + c)( b + c − a)

4

·

2 bc

2 bc

a + b + c

b + c − a

a − b + c

a + b − c

=

·

·

·

,

2

2

2

2

and since we defined s = 1 ( a

2

+ b + c), we see that

K 2 = s ( s − a)( s − b)( s − c) ,

so upon taking square roots we get

K =

s ( s − a)( s − b)( s − c) .

QED

The Area of a Triangle • Section 2.4

57

Example 2.15

Find the area of the triangle △ ABC given a = 5, b = 4, and c = 7.

C

Solution: Using Heron’s formula with s = 1 ( a + b + c) = 1 (5 + 4 + 7) = 8, the

b

2

2

= 4

a = 5

area K is given by:

A

c = 7

B

K =

s ( s − a)( s − b)( s − c)

=

8 (8 − 5)(8 − 4)(8 − 7) =

96

⇒

K = 4 6 ≈ 9.8 .

Heron’s formula is useful for theoretical purposes (e.g. in deriving other formulas). How-

ever, it is not well-suited for calculator use, exhibiting what is called numerical instability

for “extreme” triangles, as in the following example.

Example 2.16

Find the area of the triangle △ ABC given a = 1000000, b = 999999.9999979, and c = 0.0000029.

Solution: To use Heron’s formula, we need to calculate s = 1 ( a

2

+ b + c). Notice that the actual value

of a+ b+ c is 2000000.0000008, which has 14 digits. Most calculators can store 12-14 digits internally

(even if they display less), and hence may round off that value of a + b + c to 2000000. When we then

divide that rounded value for a + b + c by 2 to get s, some calculators (e.g. the TI-83 Plus) will give a

rounded down value of 1000000.

This is a problem because a = 1000000, and so we would get s − a = 0, causing Heron’s formula

to give us an area of 0 for the triangle! And this is indeed the incorrect answer that the TI-83 Plus

returns. Other calculators may give some other inaccurate answer, depending on how they store

values internally. The actual area - accurate to 15 decimal places - is K = 0.99999999999895, i.e. it

is basically 1.

The above example shows how problematic floating-point arithmetic can be.9 Luckily

there is a better formula10 for the area of a triangle when the three sides are known:

For a triangle △ ABC with sides a ≥ b ≥ c, the area is:

Area = K = 1

( a

4

+ ( b + c))( c − ( a − b))( c + ( a − b))( a + ( b − c))

(2.33)

To use this formula, sort the names of the sides so that a ≥ b ≥ c. Then perform the

operations inside the square root in the exact order in which they appear in the formula,

including the use of parentheses. Then take the square root and divide by 4. For the triangle

in Example 2.16, the above formula gives an answer of exactly K = 1 on the same TI-83 Plus

calculator that failed with Heron’s formula. What is amazing about this formula is that it

is just Heron’s formula rewritten! The use of parentheses is what forces the correct order of

operations for numerical stability.

9This is an issue even on modern computers. There is an excellent overview of this important subject in the

article What Every Computer Scientist Should Know About Floating-Point Arithmetic by D. Goldberg, available

at http://docs.sun.com/source/806-3568/ncg_goldberg.html

10Due to W. Kahan: http://www.eecs.berkeley.edu/~wkahan/Triangle.pdf

58

Chapter 2 • General Triangles

§2.4

Another formula11 for the area of a triangle given its three sides is given below:

For a triangle △ ABC with sides a ≥ b ≥ c, the area is:

2

Area = K = 1

a 2 c 2

(2.34)

2

− a 2+ c 2− b 2

2

For the triangle in Example 2.16, the above formula gives an answer of exactly K = 1 on

the same TI-83 Plus calculator that failed with Heron’s formula.

Exercises

For Exercises 1-6, find the area of the triangle △ ABC.

1. A = 70◦, b = 4, c = 12

2. a = 10, B = 95◦, c = 35

3. A = 10◦, B = 48◦, C = 122◦, c = 11

4. A = 171◦, B = 1◦, C = 8◦, b = 2

5. a = 2, b = 3, c = 4

6. a = 5, b = 6, c = 5

7. Find the area of the quadrilateral in Figure 2.4.3 below.

C

4

3.5

θ

6

B

D

2

5.5

A

Figure 2.4.3

Exercise 7

Figure 2.4.4

Exercise 8

8. Let ABCD be a quadrilateral which completely contains its two diagonals, as in Figure 2.4.4

above. Show that the area K of ABCD is equal to half the product of its diagonals and the sine of

the angle they form, i.e. K = 1 AC

2

· BD sin θ .

9. From formula (2.26) derive the following formula for the area of a triangle △ ABC:

a 2 sin B sin C

Area = K = 2 sin ( B+ C)

10. Show that the triangle area formula

Area = K = 1

( a

4

+ ( b + c))( c − ( a − b))( c + ( a − b))( a + ( b − c))

is equivalent to Heron’s formula. ( Hint: In Heron’s formula replace s by 1 ( a

2

+ b + c) . )

11. Show that the triangle area formula (2.34) is equivalent to Heron’s formula. ( Hint: Factor the

expression inside the square root. )

12. Find the angle A in Example 2.16, then use formula (2.23) to find the area. Did it work?

11Due to the Chinese mathematician Qiu Jiushao (ca. 1202-1261).

Circumscribed and Inscribed Circles • Section 2.5

59

2.5 Circumscribed and Inscribed Circles

Recall from the Law of Sines that any triangle △ ABC has a common ratio of sides to sines

of opposite angles, namely

a

b

c

=

=

.

sin A

sin B

sin C

This common ratio has a geometric meaning: it is the diameter (i.e. twice the radius) of

the unique circle in which △ ABC can be inscribed, called the circumscribed circle of the

triangle. Before proving this, we need to review some elementary geometry.

A central angle of a circle is an angle whose vertex is the center O of the circle and whose

sides (called radii) are line segments from O to two points on the circle. In Figure 2.5.1(a),

∠ O is a central angle and we say that it intercepts the arc BC.

A

A

O

O

D

B

B

B

C

C

C

(a) Central angle ∠ O

(b) Inscribed angle ∠ A

(c) ∠ A = ∠ D = 12 ∠ O

Figure 2.5.1

Types of angles in a circle

An inscribed angle of a circle is an angle whose vertex is a point A on the circle and

whose sides are line segments (called chords) from A to two other points on the circle. In

Figure 2.5.1(b), ∠ A is an inscribed angle that intercepts the arc BC. We state here without

proof12 a useful relation between inscribed and central angles:

Theorem 2.4. If an inscribed angle ∠ A and a central angle ∠ O intercept the same arc,

then ∠ A = 12 ∠ O . Thus, inscribed angles which intercept the same arc are equal.

Figure 2.5.1(c) shows two inscribed angles, ∠ A and ∠ D, which intercept the same arc BC

as the central angle ∠ O, and hence ∠ A = ∠ D = 12 ∠ O (so ∠ O = 2∠ A = 2∠ D ).

We will now prove our assertion about the common ratio in the Law of Sines:

Theorem 2.5. For any triangle △ ABC, the radius R of its circumscribed circle is given by:

a

b

c

2 R =

=

=

(2.35)

sin A

sin B

sin C

(Note: For a circle of diameter 1, this means a = sin A, b = sin B, and c = sin C.)

12For a proof, see pp. 210-211 in R.A. AVERY, Plane Geometry, Boston: Allyn & Bacon, 1950.

60

Chapter 2 • General Triangles

§2.5

To prove this, let O be the center of the circumscribed circle for a triangle △ ABC. Then

O can be either inside, outside, or on the triangle, as in Figure 2.5.2 below. In the first two

cases, draw a perpendicular line segment from O to AB at the point D.

C

C

C

b

a

c

c

b

a

2

D

2

A

B

b

O

a

c

R

R

A

B

R

R

O

R

O

R

A

c

D

c

B

2

2

(a) O inside △ ABC

(b) O outside △ ABC

(c) O on △ ABC

Figure 2.5.2

Circumscribed circle for △ ABC

The radii O A and OB have the same length R, so △ AOB is an isosceles triangle. Thus,

from elementary geometry we know that OD bisects both the angle ∠ AOB and the side AB.

So ∠ AOD = 1

. But since the inscribed angle

2 ∠ AOB and AD = c

2

∠ ACB and the central

angle ∠ AOB intercept the same arc AB, we know from Theorem 2.4 that ∠ ACB = 12 ∠ AOB.

Hence, ∠ ACB = ∠ AOD. So since C = ∠ ACB, we have

c

AD

c

c

sin C = sin ∠ AOD =

= 2 =

⇒

2 R =

,

O A

R

2 R

sin C

so by the Law of Sines the result follows if O is inside or outside △ ABC.

Now suppose that O is on △ ABC, say, on the side AB, as in Figure 2.5.2(c). Then AB is a diameter of the circle, so C = 90◦ by Thales’ Theorem. Hence, sin C = 1, and so 2 R = AB =

c = c

, and the result again follows by the Law of Sines.

QED

1 =

c

sin C

Example 2.17

Find the radius R of the circumscribed circle for the triangle △ ABC whose

B

sides are a = 3, b = 4, and c = 5.

5

Solution: We know that △ ABC is a right triangle. So as we see from Figure

3

2.5.3, sin A

O

= 3/5. Thus,

a

3

A

C

2 R

4

=

=

= 5

⇒

R = 2.5 .

sin A

3

5

Figure 2.5.3

Note that since R = 2.5, the diameter of the circle is 5, which is the same as

AB. Thus, AB must be a diameter of the circle, and so the center O of the

circle is the midpoint of AB.

Corollary 2.6. For any right triangle, the hypotenuse is a diameter of the circumscribed

circle, i.e. the center of the circle is the midpoint of the hypotenuse.

Circumscribed and Inscribed Circles • Section 2.5

61

For the right triangle in the above example, the circumscribed circle is simple to draw; its

center can be found by measuring a distance of 2.5 units from A along AB.

We need a different procedure for acute and obtuse triangles, since for an acute triangle

the center of the circumscribed circle will be inside the triangle, and it will be outside for

an obtuse triangle. Notice from the proof of Theorem 2.5 that the center O was on the

perpendicular bisector of one of the sides ( AB). Similar arguments for the other sides would

show that O is on the perpendicular bisectors for those sides:

Corollary 2.7. For any triangle, the center of its circumscribed circle is the intersection of

the perpendicular bisectors of the sides.

Recall from geometry how to create the perpendicular bisector

of a line segment: at each endpoint use a compass to draw an arc

d

d

with the same radius. Pick the radius large enough so that the

A

B

arcs intersect at two points, as in Figure 2.5.4. The line through

those two points is the perpendicular bisector of the line segment.

For the circumscribed circle of a triangle, you need the perpendic-

ular bisectors of only two of the sides; their intersection will be

the center of the circle.

Figure 2.5.4

Example 2.18

Find the radius R of the circumscribed circle for the triangle △ ABC from Example 2.6 in Section 2.2: a = 2, b = 3, and c = 4. Then draw the triangle and the circle.

Solution: In Example 2.6 we found A = 28.9◦, so 2 R = a

sin A =

2

sin 28.9◦ = 4.14, so R = 2.07 .

In Figure 2.5.5(a) we show how to draw △ ABC: use a ruler to draw the longest side AB of length c = 4, then use a compass to draw arcs of radius 3 and 2 centered at A and B, respectively. The

intersection of the arcs is the vertex C.

C

C

A

B

3

2

O

A

B

c = 4

(a) Drawing △ ABC

(b) Circumscribed circle

Figure 2.5.5

In Figure 2.5.5(b) we show how to draw the circumscribed circle: draw the perpendicular bisectors

of AB and AC; their intersection is the center O of the circle. Use a compass to draw the circle

centered at O which passes through A.

62

Chapter 2 • General Triangles

§2.5

Theorem 2.5 can be used to derive another formula for the area of a triangle:

Theorem 2.8. For a triangle △ ABC, let K be its area and let R be the radius of its circum-

scribed circle. Then

abc

abc

K =

(and hence R =

) .

(2.36)

4 R

4 K

To prove this, note that by Theorem 2.5 we have

a

b

c

a

b

c

2 R =

=

=

⇒

sin A =

, sin B =

, sin C =

.

sin A

sin B

sin C

2 R

2 R

2 R

Substitute those expressions into formula (2.26) from Section 2.4 for the area K :

a 2 sin B sin C

a 2 · b

abc

K =

=

2 R ·

c

2 R

=

QED

2 sin A

2 · a

4 R

2 R

Combining Theorem 2.8 with Heron’s formula for the area of a triangle, we get:

Corollary 2.9. For a triangle △ ABC, let s = 1 ( a

2

+ b + c). Then the radius R of its circum-

scribed circle is

abc

R =

.

(2.37)

4

s ( s − a)( s − b)( s − c)

In addition to a circumscribed circle, every triangle has an inscribed circle, i.e. a circle

to which the sides of the triangle are tangent, as in Figure 2.5.6.

C

E

F

a

b

O

r

A

B

c

D

Figure 2.5.6

Inscribed circle for △ ABC

Let r be the radius of the inscribed circle, and let D, E, and F be the points on AB, BC, and AC, respectively, at which the circle is tangent. Then OD ⊥ AB, OE ⊥ BC, and OF ⊥ AC.

Thus, △ OAD and △ OAF are equivalent triangles, since they are right triangles with the

same hypotenuse O A and with corresponding legs OD and OF of the same length r. Hence,

∠ OAD = ∠ OAF, which means that OA bisects the angle A. Similarly, OB bisects B and

OC bisects C. We have thus shown:

For any triangle, the center of its inscribed circle is the intersection of the bisectors of

the angles.

Circumscribed and Inscribed Circles • Section 2.5

63

We will use Figure 2.5.6 to find the radius r of the inscribed circle. Since O A bisects A, we see that tan 1 A

, and so r

A. Now,

2

= r

AD

= AD · tan 12

△ OAD and △ OAF are equivalent

triangles, so AD = AF. Similarly, DB = EB and FC = CE. Thus, if we let s = 1 ( a 2

+ b + c), we

see that

2 s = a + b + c = ( AD + DB) + ( CE + EB) + ( AF + FC)

= AD + EB + CE + EB + AD + CE = 2( AD + EB + CE)

s = AD + EB + CE = AD + a

AD = s − a .

Hence, r = ( s − a) tan 1 A. Similar arguments for the angles B and C give us:

2

Theorem 2.10. For any triangle △ ABC, let s = 1 ( a

2

+ b+ c). Then the radius r of its inscribed

circle is

r = ( s − a) tan 1 A

B

C .

(2.38)

2

= ( s − b) tan 12

= ( s − c) tan 12

We also see from Figure 2.5.6 that the area of the triangle △ AOB is

Area(△ AOB) = 1 base

c r .

2

× height = 12

Similarly, Area(△ BOC) = 1 a r and Area(

b r. Thus, the area K of

2

△ AOC) = 12

△ ABC is

K = Area(△ AOB) + Area(△ BOC) + Area(△ AOC) = 1 c r

a r

b r

2

+ 12

+ 12

= 1 ( a

2

+ b + c) r = sr , so by Heron’s formula we get

K

s ( s − a)( s − b)( s − c)

s ( s − a)( s − b)( s − c)

( s − a)( s − b)( s − c)

r =

=

=

=

.

s

s

s 2

s

We have thus proved the following theorem:

Theorem 2.11. For any triangle △ ABC, let s = 1 ( a

2

+ b+ c). Then the radius r of its inscribed

circle is

K

( s − a)( s − b)( s − c)

r =

=

.

(2.39)

s

s

Recall from geometry how to bisect an angle: use a compass cen-

tered at the vertex to draw an arc that intersects the sides of the

d

angle at two points. At those two points use a compass to draw an

arc with the same radius, large enough so that the two arcs inter-

d

A

sect at a point, as in Figure 2.5.7. The line through that point and

the vertex is the bisector of the angle. For the inscribed circle of a

Figure 2.5.7

triangle, you need only two angle bisectors; their intersection will be

the center of the circle.

64

Chapter 2 • General Triangles

§2.5

Example 2.19

Find the radius r of the inscribed circle for the triangle △ ABC from Example 2.6 in Section 2.2: a = 2, b = 3, and c = 4. Draw the circle.

Solution: Using Theorem 2.11 with

C

s = 1 ( a

(2

, we have

2

+ b + c) = 12 + 3 + 4) = 92

9

( s − a)( s − b)( s − c)

5

r

2 − 2

9

2 − 3

9

2 − 4

.

O

=

=

=

s

9

12

2

A

B

Figure 2.5.8 shows how to draw the inscribed circle: draw the

bisectors of A and B, then at their intersection use a compass

Figure 2.5.8

to draw a circle of radius r = 5/12 ≈ 0.645.

Exercises

For Exercises 1-6, find the radii R and r of the circumscribed and inscribed circles, respectively, of

the triangle △ ABC.

1. a = 2, b = 4, c = 5

2. a = 6, b = 8, c = 8

3. a = 5, b = 7, C = 40◦

4. A = 170◦, b = 100, c = 300

5. a = 10, b = 11, c = 20.5

6. a = 5, b = 12, c = 13

For Exercises 7 and 8, draw the triangle △ ABC and its circumscribed and inscribed circles accu-

rately, using a ruler and compass (or computer software).

7. a = 2 in, b = 4 in, c = 5 in

8. a = 5 in, b = 6 in, c = 7 in

9. For any triangle △ ABC, let s = 1 ( a

2

+ b + c). Show that

( s − b)( s − c)

( s − a)( s − c)

( s − a)( s − b)

tan 1 A

, tan 1 B

, tan 1 C

.

2

=

=

=

s ( s − a)

2

s ( s − b)

2

s ( s − c)

10. Show that for any triangle △ ABC, the radius R of its circumscribed circle is

abc

R =

.

( a + b + c)( b + c − a)( a − b + c)( a + b − c) 11. Show that for any triangle △ ABC, the radius R of its circumscribed circle and the radius r of its

inscribed circle satisfy the relation

abc

rR =

.

2 ( a + b + c)

12. Let △ ABC be an equilateral triangle whose sides are of length a.

(a) Find the exact value of the radius R of the circumscribed circle of △ ABC.

(b) Find the exact value of the radius r of the inscribed circle of △ ABC.

(c) How much larger is R than r?

(d) Show that the circumscribed and inscribed circles of △ ABC have the same center.

13. Let △ ABC be a right triangle with C = 90◦. Show that tan 1 A

.

2

=

c− b

c+ b

3 Identities

3.1 Basic Trigonometric Identities

So far we know a few relations between the trigonometric functions. For example, we know

the reciprocal relations:

1

1. csc θ =

when sin θ = 0

sin θ

1

2. sec θ =

when cos θ = 0

cos θ

1

3. cot θ =

when tan θ is defined and not 0

tan θ

1

4. sin θ =

when csc θ is defined and not 0

csc θ

1

5. cos θ =

when sec θ is defined and not 0

sec θ

1

6. tan θ =

when cot θ is defined and not 0

cot θ

Notice that each of these equations is true for all angles θ for which both sides of the equa-

tion are defined. Such equations are called identities, and in this section we will discuss

several trigonometric identities, i.e. identities involving the trigonometric functions. These

identities are often used to simplify complicated expressions or equations. For example, one

of the most useful trigonometric identities is the following:

sin θ

tan θ =

when cos θ = 0

(3.1)

cos θ

To prove this identity, pick a point ( x, y) on the terminal side of θ a distance r > 0 from the

origin, and suppose that cos θ = 0. Then x = 0 (since cos θ = x ), so by definition

r

y

sin θ

r

y

=

=

= tan θ .

cos θ

x

x

r

65

66

Chapter 3 • Identities

§3.1

Note how we proved the identity by expanding one of its sides ( sin θ ) until we got an expres-

cos θ

sion that was equal to the other side (tan θ). This is probably the most common technique

for proving identities. Taking reciprocals in the above identity gives:

cos θ

cot θ =

when sin θ = 0

(3.2)

sin θ

We will now derive one of the most important trigonometric

y

identities. Let θ be any angle with a point ( x, y) on its terminal

θ

| x|

side a distance r > 0 from the origin. By the Pythagorean The-

x

orem, r 2

0

= x 2 + y 2 (and hence r =

x 2 + y 2). For example, if θ

| y|

is in QIII as in Figure 3.1.1, then the legs of the right triangle

r

formed by the reference angle have lengths | x| and | y| (we use

( x, y)

absolute values because x and y are negative in QIII). The same

argument holds if

Figure 3.1.1

θ is in the other quadrants or on either axis.

Thus,

r 2 = | x|2 + | y|2 = x 2 + y 2 ,

so dividing both sides of the equation by r 2 (which we can do since r > 0) gives

r 2

x 2 + y 2

x 2

y 2

x 2

y 2

=

=

+

=

+

.

r 2

r 2

r 2

r 2

r

r

Since r 2

r 2 = 1, x

r = cos θ, and y

r = sin θ, we can rewrite this as:

cos2 θ + sin2 θ = 1

(3.3)

You can think of this as sort of a trigonometric variant of the Pythagorean Theorem. Note

that we use the notation sin2 θ to mean (sin θ)2, likewise for cosine and the other trigono-

metric functions. We will use the same notation for other powers besides 2.

From the above identity we can derive more identities. For example:

sin2 θ = 1 − cos2 θ

(3.4)

cos2 θ = 1 − sin2 θ

(3.5)

from which we get (after taking square roots):

sin θ = ± 1 − cos2 θ

(3.6)

cos θ = ± 1 − sin2 θ

(3.7)

Basic Trigonometric Identities • Section 3.1

67

Also, from the inequalities 0 ≤ sin2 θ = 1 − cos2 θ ≤ 1 and 0 ≤ cos2 θ = 1 − sin2 θ ≤ 1, taking

square roots gives us the following bounds on sine and cosine:

− 1 ≤ sin θ ≤ 1

(3.8)

− 1 ≤ cos θ ≤ 1

(3.9)

The above inequalities are not identities (since they are not equations), but they provide

useful checks on calculations. Recall that we derived those inequalities from the definitions

of sine and cosine in Section 1.4.

In formula (3.3), dividing both sides of the identity by cos2 θ gives

cos2 θ

sin2 θ

1

+

=

,

cos2 θ

cos2 θ

cos2 θ

so since tan θ = sin θ and sec θ

, we get:

cos θ

= 1

cos θ

1 + tan2 θ = sec2 θ

(3.10)

Likewise, dividing both sides of formula (3.3) by sin2 θ gives

cos2 θ

sin2 θ

1

+

=

,

sin2 θ

sin2 θ

sin2 θ

so since cot θ = cos θ and csc θ

, we get:

sin θ

= 1

sin θ

cot2 θ + 1 = csc2 θ

(3.11)

Example 3.1

Simplify cos2 θ tan2 θ .

Solution: We can use formula (3.1) to simplify:

sin2 θ

cos2 θ tan2 θ = cos2 θ · cos2 θ

= sin2 θ

Example 3.2

Simplify 5 sin2 θ + 4cos2 θ .

Solution: We can use formula (3.5) to simplify:

5 sin2 θ + 4cos2 θ = 5sin2 θ + 4 1 − sin2 θ

= 5sin2 θ + 4 − 4sin2 θ

= sin2 θ + 4

68

Chapter 3 • Identities

§3.1

Example 3.3

Prove that tan θ + cot θ = sec θ csc θ .

Solution: We will expand the left side and show that it equals the right side:

sin θ

cos θ

tan θ + cot θ =

+

(by (3.1) and (3.2))

cos θ

sin θ

sin θ

sin θ

cos θ

cos θ

=

·

+

·

(multiply both fractions by 1)

cos θ

sin θ

sin θ

cos θ

sin2 θ + cos2 θ

=

(after getting a common denominator)

cos θ sin θ

1

=

(by (3.3))

cos θ sin θ

1

1

=

·

cos θ

sin θ

= sec θ csc θ

In the above example, how did we know to expand the left side instead of the right side?

In general, though this technique does not always work, the more complicated side of the

identity is likely to be easier to expand. The reason is that, by its complexity, there will be

more things that you can do with that expression. For example, if you were asked to prove

that

sec θ − sin θ tan θ = cos θ ,

there would not be much that you could do with the right side of that identity; it consists of

a single term (cos θ) that offers no obvious means of expansion.

Example 3.4

1 + cot2 θ

Prove that

= csc θ cot θ .

sec θ

Solution: Of the two sides, the left side looks more complicated, so we will expand that:

1 + cot2 θ

csc2 θ

=

(by (3.11))

sec θ

sec θ

1

csc θ · sin θ

=

1

cos θ

cos θ

= csc θ · sin θ

= csc θ cot θ

(by (3.2))

Basic Trigonometric Identities • Section 3.1

69

Example 3.5

tan2 θ + 2

Prove that

= 1 + cos2 θ .

1 + tan2 θ

Solution: Expand the left side:

tan2 θ + 2

tan2 θ + 1 + 1

=

1 + tan2 θ

1 + tan2 θ

sec2 θ + 1

=

(by (3.10))

sec2 θ

sec2 θ

1

=

+

sec2 θ

sec2 θ

= 1 + cos2 θ

When trying to prove an identity where at least one side is a ratio of expressions, cross-

multiplying can be an effective technique:

a

c

=

if and only if

ad = bc

b

d

Example 3.6

1 + sin θ

cos θ

Prove that

=

.

cos θ

1 − sin θ

Solution: Cross-multiply and reduce both sides until it is clear that they are equal:

(1 + sin θ)(1 − sin θ) = cos θ · cos θ

1 − sin2 θ = cos2 θ

By (3.5) both sides of the last equation are indeed equal. Thus, the original identity holds.

Example 3.7

Suppose that a cos θ = b and c sin θ = d for some angle θ and some constants a, b, c, and d. Show that a 2 c 2 = b 2 c 2 + a 2 d 2.

Solution: Multiply both sides of the first equation by c and the second equation by a:

ac cos θ = bc

ac sin θ = ad

Now square each of the above equations then add them together to get:

( ac cos θ)2 + ( ac sin θ)2 = ( bc)2 + ( ad)2

( ac)2 cos2 θ + sin2 θ = b 2 c 2 + a 2 d 2

a 2 c 2 = b 2 c 2 + a 2 d 2

(by (3.3))

Notice how θ does not appear in our final result. The trick was to get a common coefficient ( ac) for

cos θ and sin θ so that we could use cos2 θ + sin2 θ = 1. This is a common technique for eliminating

trigonometric functions from systems of equations.

70

Chapter 3 • Identities

§3.1

Exercises

1. We showed that sin θ = ± 1 − cos2 θ for all θ. Give an example of an angle θ such that

sin θ = − 1 − cos2 θ .

2. We showed that cos θ = ± 1 − sin2 θ for all θ. Give an example of an angle θ such that

cos θ = − 1 − sin2 θ .

3. Suppose that you are given a system of two equations of the following form:1

A cos φ = B ν 1 − Bν 2 cos θ

A sin φ = B ν 2 sin θ .

Show that A 2 = B 2 ν 21 + ν 22 − 2 ν 1 ν 2 cos θ .

For Exercises 4-16, prove the given identity.

4. cos θ tan θ = sin θ

5. sin θ cot θ = cos θ

tan θ

csc θ

6.

= tan2 θ

7.

= csc2 θ

cot θ

sin θ

cos2 θ

1 − 2 cos2 θ

8.

= 1 − sin θ

9.

= tan θ − cot θ

1 + sin θ

sin θ cos θ

10. sin4 θ − cos4 θ = sin2 θ − cos2 θ

11. cos4 θ − sin4 θ = 1 − 2 sin2 θ

1 − tan θ

cot θ − 1

tan θ + tan φ

12.

=

13.

= tan θ tan φ

1 + tan θ

cot θ + 1

cot θ + cot φ

sin2 θ

1 − tan2 θ

14.

= tan2 θ

15.

= 1 − sec2 θ

1 − sin2 θ

1 − cot2 θ

tan θ

16. sin θ = ±

( Hint: Solve for sin2 θ in Exercise 14. )

1 + tan2 θ

17. Sometimes identities can be proved by geometrical methods. For ex-

y

(1, y)

ample, to prove the identity in Exercise 16, draw an acute angle θ in QI

y

and pick the point (1, y) on its terminal side, as in Figure 3.1.2. What

θ

must y equal? Use that to prove the identity for acute θ. Explain the

x

0

adjustment(s) you would need to make in Figure 3.1.2 to prove the iden-

1

tity for θ in the other quadrants. Does the identity hold if θ is on either

axis?

Figure 3.1.2

18. Similar to Exercise 16 , find an expression for cos θ solely in terms of tan θ.

19. Find an expression for tan θ solely in terms of sin θ, and one solely in terms of cos θ.

20. Suppose that a point with coordinates ( x, y) = ( a (cos ψ − ǫ), a 1 − ǫ 2 sin ψ) is a distance r > 0

from the origin, where a > 0 and 0 < ǫ < 1. Use r 2 = x 2 + y 2 to show that r = a (1 − ǫ cos ψ) .

(Note: These coordinates arise in the study of elliptical orbits of planets.)

21. Show that each trigonometric function can be put in terms of the sine function.

1These types of equations arise in physics, e.g. in the study of photon-electron collisions. See pp. 95-97 in W.

RINDLER, Special Relativity, Edinburgh: Oliver and Boyd, LTD., 1960.

Sum and Difference Formulas • Section 3.2

71

3.2 Sum and Difference Formulas

We will now derive identities for the trigonometric functions of the sum and difference of

two angles. For the sum of any two angles A and B, we have the addition formulas:

sin ( A + B) = sin A cos B + cos A sin B

(3.12)

cos ( A + B) = cos A cos B − sin A sin B

(3.13)

To prove these, first assume that A and B are acute angles. Then A + B is either acute or

obtuse, as in Figure 3.2.1. Note in both cases that ∠ QPR = A, since

∠ QPR = ∠ QPO − ∠ OP M = (90◦ − B) − (90◦ − ( A + B)) = A in Figure 3.2.1(a), and

∠ QPR = ∠ QPO + ∠ OP M = (90◦ − B) + (90◦ − (180◦ − ( A + B))) = A in Figure 3.2.1(b).

P

P

A

A

Q

R

A + B

R

Q

A + B

B

B

A

A

O

N

N

M

M

O

(a) A + B acute

(b) A + B obtuse

Figure 3.2.1

sin ( A + B) and cos ( A + B) for acute A and B

Thus,

MP

MR + RP

NQ + RP

NQ

RP

sin ( A + B) =

=

=

=

+

OP

OP

OP

OP

OP

NQ

OQ

RP

PQ

=

·

+

·

OQ

OP

PQ

OP

= sin A cos B + cos A sin B ,

(3.14)

72

Chapter 3 • Identities

§3.2

and

OM

ON − MN

ON − RQ

ON

RQ

cos ( A + B) =

=

=

=

−

OP

OP

OP

OP

OP

ON

OQ

RQ

PQ

=

·

+

·

OQ

OP

PQ

OP

= cos A cos B − sin A sin B .

(3.15)

So we have proved the identities for acute angles A and B. It is simple to verify that they

hold in the special case of A = B = 0◦. For general angles, we will need to use the relations

we derived in Section 1.5 which involve adding or subtracting 90◦:

sin ( θ + 90◦) =

cos θ

sin ( θ − 90◦) = −cos θ

cos ( θ + 90◦) = −sin θ

cos ( θ − 90◦) =

sin θ

These will be useful because any angle can be written as the sum of an acute angle (or

0◦ ) and integer multiples of ±90◦. For example, 155◦ = 65◦ + 90◦, 222◦ = 42◦ + 2(90◦), −77◦ =

13◦ − 90◦, etc. So if we can prove that the identities hold when adding or subtracting 90◦

to or from either A or B, respectively, where A and B are acute or 0◦, then the identities

will also hold when repeatedly adding or subtracting 90◦, and hence will hold for all angles.

Replacing A by A + 90◦ and using the relations for adding 90◦ gives

sin (( A + 90◦) + B) = sin (( A + B) + 90◦) = cos ( A + B) ,

= cos A cos B − sin A sin B (by equation (3.15))

= sin ( A + 90◦) cos B + cos ( A + 90◦) sin B ,

so the identity holds for A + 90◦ and B (and, similarly, for A and B + 90◦). Likewise,

sin (( A − 90◦) + B) = sin (( A + B) − 90◦) = −cos ( A + B) ,

= −(cos A cos B − sin A sin B)

= (−cos A) cos B + sin A sin B

= sin ( A − 90◦) cos B + cos ( A − 90◦) sin B ,

so the identity holds for A −90◦ and B (and, similarly, for A and B +90◦). Thus, the addition

formula (3.12) for sine holds for all A and B. A similar argument shows that the addition

formula (3.13) for cosine is true for all A and B.

QED

Replacing B by − B in the addition formulas and using the relations sin (− θ) = −sin θ and

cos (− θ) = cos θ from Section 1.5 gives us the subtraction formulas:

sin ( A − B) = sin A cos B − cos A sin B

(3.16)

cos ( A − B) = cos A cos B + sin A sin B

(3.17)

Sum and Difference Formulas • Section 3.2

73

Using the identity tan θ = sin θ , and the addition formulas for sine and cosine, we can

cos θ

derive the addition formula for tangent:

sin ( A + B)

tan ( A + B) = cos ( A + B)

sin A cos B + cos A sin B

= cos A cos B − sin A sin B

sin A cos B

cos A sin B

+

cos A cos B

cos A cos B

=

(divide top and bottom by cos A cos B)

cos A cos B

sin A sin B

−

cos A cos B

cos A cos B

sin A

cos B

cos A

sin B

·

+

·

cos A

cos B

cos A

cos B

tan A + tan B

=

=

sin A

sin B

1 − tan A tan B

1 −

·

cos A

cos B

This, combined with replacing B by − B and using the relation tan (− θ) = −tan θ, gives us

the addition and subtraction formulas for tangent:

tan A + tan B

tan ( A + B) =

(3.18)

1 − tan A tan B

tan A − tan B

tan ( A − B) =

(3.19)

1 + tan A tan B

Example 3.8

Given angles A and B such that sin A = 4 , cos A

, sin B

, and cos B

, find the exact values

5

= 35

= 12

13

= 5

13

of sin ( A + B), cos ( A + B), and tan ( A + B).

Solution: Using the addition formula for sine, we get:

sin ( A + B) = sin A cos B + cos A sin B

4

5

3

12

56

=

·

+

·

⇒

sin ( A + B) =

5

13

5

13

65

Using the addition formula for cosine, we get:

cos ( A + B) = cos A cos B − sin A sin B

3

5

4

12

33

=

·

−

·

⇒

cos ( A + B) = −

5

13

5

13

65

Instead of using the addition formula for tangent, we can use the results above:

56

sin ( A + B)

56

tan ( A + B) =

= 65

⇒

tan ( A + B) = −

cos ( A + B)

− 33

33

65

74

Chapter 3 • Identities

§3.2

Example 3.9

Prove the following identity:

sin ( A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C

Solution: Treat A + B + C as ( A + B) + C and use the addition formulas three times: sin ( A + B + C) = sin (( A + B) + C)

= sin ( A + B) cos C + cos ( A + B) sin C

= (sin A cos B + cos A sin B) cos C + (cos A cos B − sin A sin B) sin C

= sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C

Example 3.10

For any triangle △ ABC, show that tan A + tan B + tan C = tan A tan B tan C.

Solution: Note that this is not an identity which holds for all angles; since A, B, and C are the angles of a triangle, it holds when A, B, C > 0◦ and A + B + C = 180◦. So using C = 180◦ − ( A + B) and the relation tan (180◦ − θ) = −tan θ from Section 1.5, we get:

tan A + tan B + tan C = tan A + tan B + tan (180◦ − ( A + B))

= tan A + tan B − tan ( A + B)

tan A + tan B

= tan A + tan B − 1−tan A tan B

1

= (tan A + tan B) 1 − 1−tan A tan B

1 − tan A tan B

1

= (tan A + tan B)

−

1 − tan A tan B

1 − tan A tan B

−tan A tan B

= (tan A + tan B) · 1−tan A tan B

tan A + tan B

= tan A tan B · − 1−tan A tan B

= tan A tan B · (−tan ( A + B))

= tan A tan B · (tan (180◦ − ( A + B)))

= tan A tan B tan C

Example 3.11

Let A, B, C, and D be positive angles such that A + B + C + D = 180◦. Show that2

sin A sin B + sin C sin D = sin ( A + C) sin ( B + C) .

Solution: It may be tempting to expand the right side, since it appears more complicated. However,

notice that the right side has no D term. So instead, we will expand the left side, since we can

eliminate the D term on that side by using D = 180◦ − ( A + B + C) and the relation

sin (180◦ − ( A + B + C)) = sin ( A + B + C).

2This is the “trigonometric form” of Ptolemy’s Theorem, which says that a quadrilateral can be inscribed in a

circle if and only if the sum of the products of its opposite sides equals the product of its diagonals.

Sum and Difference Formulas • Section 3.2

75

So since sin D = sin ( A + B + C), we get

sin A sin B + sin C sin D = sin A sin B + sin C sin ( A + B + C) , so by Example 3.9 we get

= sin A sin B + sin C (sin A cos B cos C + cos A sin B cos C

+ cos A cos B sin C − sin A sin B sin C)

= sin A sin B + sin C sin A cos B cos C + sin C cos A sin B cos C

+ sin C cos A cos B sin C − sin C sin A sin B sin C .

It may not be immediately obvious where to go from here, but it is not completely guesswork. We

need to end up with sin ( A + C) sin ( B + C), and we know that sin ( B + C) = sin B cos C + cos B sin C.

There are two terms involving cos B sin C, so group them together to get

sin A sin B + sin C sin D = sin A sin B − sin C sin A sin B sin C + sin C cos A sin B cos C

+ cos B sin C (sin A cos C + cos A sin C)

= sin A sin B (1 − sin2 C) + sin C cos A sin B cos C

+ cos B sin C sin ( A + C)

= sin A sin B cos2 C + sin C cos A sin B cos C

+ cos B sin C sin ( A + C) .

We now have two terms involving sin B cos C, which we can factor out:

sin A sin B + sin C sin D = sin B cos C (sin A cos C + cos A sin C )

+ cos B sin C sin ( A + C)

= sin B cos C sin ( A + C) + cos B sin C sin ( A + C)

= sin ( A + C) (sin B cos C + cos B sin C)

= sin ( A + C) sin ( B + C)

Example 3.12

In the study of the propagation of electromagnetic waves, Snell’s law gives the relation

n 1 sin θ 1 = n 2 sin θ 2 ,

(3.20)

where θ 1 is the angle of incidence at which a wave strikes the planar boundary between two mediums,

θ 2 is the angle of transmission of the wave through the new medium, and n 1 and n 2 are the indexes

of refraction of the two mediums. The quantity

n 1 cos θ 1 − n 2 cos θ 2

r 1 2 s =

(3.21)

n 1 cos θ 1 + n 2 cos θ 2

is called the Fresnel coefficient for normal incidence reflection of the wave for s-polarization. Show

that this can be written as:

sin ( θ 2 − θ 1)

r 1 2 s = sin ( θ 2 + θ 1)

76

Chapter 3 • Identities

§3.2

Solution: Multiply the top and bottom of r 1 2 s by sin θ 1 sin θ 2 to get:

n 1 cos θ 1 − n 2 cos θ 2

sin θ 1 sin θ 2

r 1 2 s =

·

n 1 cos θ 1 + n 2 cos θ 2

sin θ 1 sin θ 2

( n 1 sin θ 1) sin θ 2 cos θ 1 − ( n 2 sin θ 2) cos θ 2 sin θ 1

= ( n 1 sin θ 1) sin θ 2 cos θ 1 + ( n 2 sin θ 2) cos θ 2 sin θ 1

( n 1 sin θ 1) sin θ 2 cos θ 1 − ( n 1 sin θ 1) cos θ 2 sin θ 1

=

(by Snell’s law)

( n 1 sin θ 1) sin θ 2 cos θ 1 + ( n 1 sin θ 1) cos θ 2 sin θ 1

sin θ 2 cos θ 1 − cos θ 2 sin θ 1

= sin θ 2 cos θ 1 + cos θ 2 sin θ 1

sin ( θ 2 − θ 1)

= sin ( θ 2 + θ 1)

The last two examples demonstrate an important aspect of how identities are used in

practice: recognizing terms which are part of known identities, so that they can be factored

out. This is a common technique.

Exercises

1. Verify the addition formulas (3.12) and (3.13) for A = B = 0◦.

For Exercises 2 and 3, find the exact values of sin ( A + B), cos ( A + B), and tan ( A + B).

2. sin A = 8 , cos A

, sin B

, cos B

3. sin A

, cos A

, sin B

, cos B

17

= 15

17

= 24

25

= 7

25

= 40

41

= 9

41

= 20

29

= 21

29

4. Use 75◦ = 45◦ + 30◦ to find the exact value of sin 75◦.

5. Use 15◦ = 45◦ − 30◦ to find the exact value of tan 15◦.

6. Prove the identity sin θ + cos θ = 2 sin ( θ + 45◦) . Explain why this shows that

− 2 ≤ sin θ + cos θ ≤

2

for all angles θ. For which θ between 0◦ and 360◦ would sin θ + cos θ be the largest?

For Exercises 7-14, prove the given identity.

7. cos ( A + B + C) = cos A cos B cos C − cos A sin B sin C − sin A cos B sin C − sin A sin B cos C

tan A + tan B + tan C − tan A tan B tan C

8. tan ( A + B + C) = 1 − tan B tan C − tan A tan C − tan A tan B

cot A cot B − 1

cot A cot B + 1

9. cot ( A + B) =

10. cot ( A − B) =

cot A + cot B

cot B − cot A

Sum and Difference Formulas • Section 3.2

77

1 + tan θ

cos ( A + B)

11. tan ( θ + 45◦) =

12.

= cot A − tan B

1 − tan θ

sin A cos B

sin ( A + B)

sin ( A − B)

cot B − cot A

13. cot A + cot B =

14.

=

sin A sin B

sin ( A + B)

cot B + cot A

15. Generalize Exercise 6: For any a and b, − a 2 + b 2 ≤ a sin θ + b cos θ ≤

a 2 + b 2 for all θ.

16. Continuing Example 3.12, use Snell’s law to show that the s-polarization transmission Fresnel

coefficient

2 n 1 cos θ 1

t 1 2 s =

(3.22)

n 1 cos θ 1 + n 2 cos θ 2

can be written as:

2 cos θ 1 sin θ 2

t 1 2 s =

sin ( θ 2 + θ 1)

y

17. Suppose that two lines with slopes m 1 and m 2, respectively,

intersect at an angle θ and are not perpendicular (i.e. θ =

y = m 2 x + b 2

90◦), as in the figure on the right. Show that

y = m 1 x + b 1

θ

m 1 − m 2

tan θ =

.

x

1 + m 1 m 2

0

( Hint: Use Example 1.26 from Section 1.5. )

18. Use Exercise 17 to find the angle between the lines y = 2 x + 3 and y = −5 x − 4.

19. For any triangle △ ABC, show that cot A cot B + cot B cot C + cot C cot A = 1.

( Hint: Use Exercise 9 and C = 180◦ − ( A + B) . )

20. For any positive angles A, B, and C such that A + B + C = 90◦, show that

tan A tan B + tan B tan C + tan C tan A = 1 .

21. Prove the identity sin ( A + B) cos B − cos ( A + B) sin B = sin A.

Note that the right side depends only on A, while the left side depends on both A and B.

22. A line segment of length r > 0 from the origin to the point ( x, y) makes an angle α with the

positive x-axis, so that ( x, y) = ( r cos α, r sin α), as in the figure below. What are the endpoint’s new coordinates ( x′, y′) after a counterclockwise rotation by an angle β ? Your answer should be in

terms of r, α, and β.

y

( x, y) = ( r cos α, r sin α)

r

( x′, y′)

r

β

α

x

0

78

Chapter 3 • Identities

§3.3

3.3 Double-Angle and Half-Angle Formulas

A special case of the addition formulas is when the two angles being added are equal, result-

ing in the double-angle formulas:

sin 2 θ = 2 sin θ cos θ

(3.23)

cos 2 θ = cos2 θ − sin2 θ

(3.24)

2 tan θ

tan 2 θ =

(3.25)

1 − tan2 θ

To derive the sine double-angle formula, we see that

sin 2 θ = sin ( θ + θ) = sin θ cos θ + cos θ sin θ = 2 sin θ cos θ .

Likewise, for the cosine double-angle formula, we have

cos 2 θ = cos ( θ + θ) = cos θ cos θ − sin θ sin θ = cos2 θ − sin2 θ , and for the tangent we get

tan θ + tan θ

2 tan θ

tan 2 θ = tan ( θ + θ) =

=

1 − tan θ tan θ

1 − tan2 θ

Using the identities sin2 θ = 1 − cos2 θ and cos2 θ = 1 − sin2 θ, we get the following useful

alternate forms for the cosine double-angle formula:

cos 2 θ = 2 cos2 θ − 1

(3.26)

= 1 − 2 sin2 θ

(3.27)

Example 3.13

Prove that sin 3 θ = 3 sin θ − 4 sin3 θ .

Solution: Using 3 θ = 2 θ + θ, the addition formula for sine, and the double-angle formulas (3.23) and

(3.27), we get:

sin 3 θ = sin (2 θ + θ)

= sin 2 θ cos θ + cos 2 θ sin θ

= (2 sin θ cos θ) cos θ + (1 − 2 sin2 θ) sin θ

= 2 sin θ cos2 θ + sin θ − 2 sin3 θ

= 2 sin θ (1 − sin2 θ) + sin θ − 2 sin3 θ

= 3 sin θ − 4 sin3 θ

Double-Angle and Half-Angle Formulas • Section 3.3

79

Example 3.14

4 tan z (1 − tan2 z)

Prove that sin 4 z =

.

(1 + tan2 z)2

Solution: Expand the right side and use 1 + tan2 z = sec2 z :

sin z

cos2 z

sin2 z

4 ·

·

−

4 tan z (1 − tan2 z)

cos z

cos2 z

cos2 z

=

(1 + tan2 z)2

(sec2 z)2

sin z

cos 2 z

4 ·

·

cos z

cos2 z

=

(by formula (3.24))

1

2

cos2 z

= (4 sin z cos 2 z) cos z

= 2 (2 sin z cos z) cos 2 z

= 2 sin 2 z cos 2 z

(by formula (3.23))

= sin 4 z

(by formula (3.23) with θ replaced by 2 z)

Note: Perhaps surprisingly, this seemingly obscure identity has found a use in physics, in the deriva-

tion of a solution of the sine-Gordon equation in the theory of nonlinear waves.3

Closely related to the double-angle formulas are the half-angle formulas:

1 − cos θ

sin2 1 θ

(3.28)

2

=

2

1 + cos θ

cos2 1 θ

(3.29)

2

=

2

1 − cos θ

tan2 1 θ

(3.30)

2

= 1 + cos θ

These formulas are just the double-angle formulas rewritten with θ replaced by 1 θ:

2

1 − cos 2 θ

1 − cos 2( 1 θ)

1 − cos θ

cos 2 θ = 1 − 2 sin2 θ ⇒ sin2 θ =

⇒ sin2 1 θ =

2

=

2

2

2

2

1 + cos 2 θ

1 + cos 2( 1 θ)

1 + cos θ

cos 2 θ = 2 cos2 θ − 1 ⇒ cos2 θ =

⇒ cos2 1 θ =

2

=

2

2

2

2

The tangent half-angle formula then follows easily:

2

sin 1 θ

sin2 1 θ

1 − cos θ

1 − cos θ

tan2 1 θ

2

2

2

2

=

=

=

=

cos 1 θ

cos2 1 θ

1 + cos θ

1 + cos θ

2

2

2

The half-angle formulas are often used (e.g. in calculus) to replace a squared trigonometric

function by a nonsquared function, especially when 2 θ is used instead of θ.

3See p.331 in L.A. OSTROVSKY AND A.I.POTAPOV, Modulated Waves: Theory and Applications, Baltimore:

The Johns Hopkins University Press, 1999.

80

Chapter 3 • Identities

§3.3

By taking square roots, we can write the above formulas in an alternate form:

1 − cos θ

sin 1 θ

(3.31)

2

= ±

2

1 + cos θ

cos 1 θ

(3.32)

2

= ±

2

1 − cos θ

tan 1 θ

(3.33)

2

= ±

1 + cos θ

In the above form, the sign in front of the square root is determined by the quadrant in

which the angle 1 θ is located. For example, if θ

θ

2

= 300◦ then 12 = 150◦ is in QII. So in this

case cos 1 θ

θ

.

2

< 0 and hence we would have cos 12 = −

1 + cos θ

2

In formula (3.33), multiplying the numerator and denominator inside the square root by

(1 − cos θ) gives

1 − cos θ 1 − cos θ

(1 − cos θ)2

(1 − cos θ)2

1 − cos θ

tan 1 θ

.

2

= ±

·

= ±

= ±

= ±

1 + cos θ 1 − cos θ

1 − cos2 θ

sin2 θ

sin θ

But 1 − cos θ ≥ 0, and it turns out (see Exercise 10) that tan 1 θ and sin θ always have the 2

same sign. Thus, the minus sign in front of the last expression is not possible (since that

would switch the signs of tan 1 θ and sin θ), so we have:

2

1 − cos θ

tan 1 θ

(3.34)

2

=

sin θ

Multiplying the numerator and denominator in formula (3.34) by 1 + cos θ gives

1 − cos θ

1 + cos θ

1 − cos2 θ

sin2 θ

tan 1 θ

,

2

=

·

=

=

sin θ

1 + cos θ

sin θ (1 + cos θ)

sin θ (1 + cos θ)

so we also get:

sin θ

tan 1 θ

(3.35)

2

= 1 + cos θ

Taking reciprocals in formulas (3.34) and (3.35) gives:

sin θ

1 + cos θ

cot 1 θ

(3.36)

2

=

=

1 − cos θ

sin θ

Double-Angle and Half-Angle Formulas • Section 3.3

81

Example 3.15

2 sec θ

Prove the identity sec2 1 θ

.

2

= sec θ + 1

Solution: Since secant is the reciprocal of cosine, taking the reciprocal of formula (3.29) for cos2 1 θ

2

gives us

2

2

sec θ

2 sec θ

sec2 1 θ

.

2

=

=

·

=

1 + cos θ

1 + cos θ

sec θ

sec θ + 1

Exercises

For Exercises 1-8, prove the given identity.

1. cos 3 θ = 4 cos3 θ − 3 cos θ

2. tan 1 θ

2

= csc θ − cot θ

sin 2 θ

cos 2 θ

sin 3 θ

cos 3 θ

3.

−

= sec θ

4.

−

= 2

sin θ

cos θ

sin θ

cos θ

2

3 tan θ

5. tan 2 θ =

− tan3 θ

6. tan 3 θ

cot θ − tan θ

=

1 − 3 tan2 θ

tan θ − sin θ

cos2 ψ

1

7. tan2 1 θ

+ cos 2 ψ

2

=

8.

tan θ

=

+ sin θ

cos2 θ

1 + cos 2 θ

9. Some trigonometry textbooks used to claim incorrectly that sin θ + cos θ =

1 + sin 2 θ was

an identity. Give an example of a specific angle θ that would make that equation false. Is

sin θ + cos θ = ± 1 + sin 2 θ an identity? Justify your answer.

10. Fill out the rest of the table below for the angles 0◦ < θ < 720◦ in increments of 90◦, showing θ,

1 θ, and the signs (

θ.

2

+ or −) of sin θ and tan 12

θ

1 θ

sin θ

tan 1 θ

θ

1 θ

sin θ

tan 1 θ

2

2

2

2

0◦ − 90◦

0◦ − 45◦

+

+

360◦ − 450◦

180◦ − 225◦

90◦ − 180◦

45◦ − 90◦

450◦ − 540◦

225◦ − 270◦

180◦ − 270◦

90◦ − 135◦

540◦ − 630◦

270◦ − 315◦

270◦ − 360◦

135◦ − 180◦

630◦ − 720◦

315◦ − 360◦

11. In general, what is the largest value that sin θ cos θ can take? Justify your answer.

For Exercises 12-17, prove the given identity for any right triangle △ ABC with C = 90◦.

12. sin ( A − B) = cos 2 B

13. cos ( A − B) = sin 2 A

2 ab

b 2

14. sin 2 A

− a 2

=

c 2

15. cos 2 A =

c 2

2 ab

c − b

a

16. tan 2 A =

17. tan 1 A =

=

b 2 − a 2

2

a

c + b

18. Continuing Exercise 20 from Section 3.1, it can be shown that

r (1 − cos θ) = a (1 + ǫ)(1 − cos ψ) , and

r (1 + cos θ) = a (1 − ǫ)(1 + cos ψ) ,

where θ and ψ are always in the same quadrant. Show that tan 1 θ

tan 1 ψ .

2

=

1 + ǫ

1 − ǫ

2

82

Chapter 3 • Identities

§3.4

3.4 Other Identities

Though the identities in this section fall under the category of “other”, they are perhaps

(along with cos2 θ +sin2 θ = 1) the most widely used identities in practice. It is very common

to encounter terms such as sin A + sin B or sin A cos B in calculations, so we will now

derive identities for those expressions. First, we have what are often called the product-to-

sum formulas:

sin A cos B =

1 (sin ( A

2

+ B) + sin ( A − B))

(3.37)

cos A sin B =

1 (sin ( A

2

+ B) − sin ( A − B))

(3.38)

cos A cos B =

1 (cos ( A

2

+ B) + cos ( A − B))

(3.39)

sin A sin B = − 1 (cos ( A

2

+ B) − cos ( A − B))

(3.40)

We will prove the first formula; the proofs of the others are similar (see Exercises 1-3). We

see that

sin ( A + B) + sin ( A − B) = (sin A cos B + ✭✭✭✭✭✭

cos A sin B) + (sin A cos B − ✭✭✭✭✭✭

cos A sin B)

= 2 sin A cos B ,

so formula (3.37) follows upon dividing both sides by 2. Notice how in each of the above

identities a product (e.g. sin A cos B) of trigonometric functions is shown to be equivalent to

a sum (e.g. 1 (sin ( A

2

+ B) + sin ( A− B))) of such functions. We can go in the opposite direction,

with the sum-to-product formulas:

sin A + sin B =

2 sin 1 ( A

( A

2

+ B) cos 12

− B)

(3.41)

sin A − sin B =

2 cos 1 ( A

( A

2

+ B) sin 12

− B)

(3.42)

cos A + cos B =

2 cos 1 ( A

( A

2

+ B) cos 12

− B)

(3.43)

cos A − cos B = −2 sin 1 ( A

( A

2

+ B) sin 12

− B)

(3.44)

These formulas are just the product-to-sum formulas rewritten by using some clever sub-

stitutions: let x = 1 ( A

( A

2

+ B) and y = 12

− B). Then x + y = A and x − y = B. For example, to

derive formula (3.43), make the above substitutions in formula (3.39) to get

cos A + cos B = cos ( x + y) + cos ( x − y)

= 2 · 1 (cos ( x

2

+ y) + cos ( x − y))

= 2 cos x cos y

(by formula (3.39))

= 2 cos 1 ( A

( A

2

+ B) cos 12

− B) .

The proofs of the other sum-to-product formulas are similar (see Exercises 4-6).

Other Identities • Section 3.4

83

Example 3.16

We are now in a position to prove Mollweide’s equations, which we introduced in Section 2.3: For any

triangle △ ABC,

a − b

sin 1 ( A − B)

a + b

cos 1 ( A − B)

=

2

and

=

2

.

c

cos 1 C

c

sin 1 C

2

2

First, since C = 2 · 1 C, by the double-angle formula we have sin C

C cos 1 C. Thus,

2

= 2 sin 12

2

a − b

a

b

sin A

sin B

=

−

=

−

(by the Law of Sines)

c

c

c

sin C

sin C

sin A − sin B

sin A − sin B

=

=

sin C

2 sin 1 C cos 1 C

2

2

2 cos 1 ( A + B) sin 1 ( A − B)

=

2

2

(by formula (3.42))

2 sin 1 C cos 1 C

2

2

cos 1 (180◦ − C) sin 1 ( A − B)

=

2

2

(since A + B = 180◦ − C)

sin 1 C cos 1 C

2

2

✘✘✘✘✘✘

cos (90◦ − 1 C) sin 1 ( A − B)

=

2

2

✟✟✟

sin 1 C cos 1 C

2

2

sin 1 ( A − B)

=

2

(since cos (90◦ − 1 C) = sin 1 C) .

cos 1 C

2

2

2

This proves the first equation. The proof of the other equation is similar (see Exercise 7).

Example 3.17

Using Mollweide’s equations, we can prove the Law of Tangents: For any triangle △ ABC,

a − b

tan 1 ( A − B)

b − c

tan 1 ( B − C)

c − a

tan 1 ( C − A)

=

2

,

=

2

,

=

2

.

a + b

tan 1 ( A

b + c

tan 1 ( B

c + a

tan 1 ( C

2

+ B)

2

+ C)

2

+ A)

We need only prove the first equation; the other two are obtained by cycling through the letters. We

see that

sin 1 ( A − B)

a − b

2

a − b

c

cos 1 C

=

=

2

(by Mollweide’s equations)

a + b

a + b

cos 1 ( A

2

− B)

c

sin 1 C

2

sin 1 ( A − B)

sin 1 C

=

2

·

2

cos 1 ( A

cos 1 C

2

− B)

2

= tan 1 ( A

C

( A

( A

2

− B) · tan 12

= tan 12

− B) · tan (90◦ − 12

+ B)) (since C = 180◦ − ( A + B))

= tan 1 ( A

( A

( A

( A

2

− B) · cot 12

+ B) (since tan (90◦ − 12

+ B)) = cot 12

+ B), see Section 1.5)

tan 1 ( A − B)

=

2

.

QED

tan 1 ( A

2

+ B)

84

Chapter 3 • Identities

§3.4

Example 3.18

For any triangle △ ABC, show that

cos A + cos B + cos C = 1 + 4 sin 1 A sin 1 B sin 1 C .

2

2

2

Solution: Since cos ( A + B + C) = cos 180◦ = −1, we can rewrite the left side as

cos A + cos B + cos C = 1 + (cos ( A + B + C) + cos C) + (cos A + cos B) , so by formula (3.43)

= 1 + 2 cos 1 ( A

( A

( A

( A

2

+ B + 2 C) cos 12

+ B) + 2 cos 12

+ B) cos 12

− B)

= 1 + 2 cos 1 ( A

( A

( A

2

+ B) cos 12

+ B + 2 C) + cos 12

− B) , so

= 1 + 2 cos 1 ( A

( A

( B

2

+ B) · 2 cos 12

+ C) cos 12

+ C) by formula (3.43),

since 1 1 ( A

( A

( A

1 ( A

( A

( B

2

2

+ B + 2 C) + 12

− B) = 12

+ C) and 12 2

+ B + 2 C) − 12

− B) = 12

+ C). Thus,

cos A + cos B + cos C = 1 + 4 cos (90◦ − 1 C) cos (90◦

B) cos (90◦

A)

2

− 12

− 12

= 1 + 4 sin 1 C sin 1 B sin 1 A , so rearranging the order gives

2

2

2

= 1 + 4 sin 1 A sin 1 B sin 1 C .

2

2

2

Example 3.19

For any triangle △ ABC, show that sin 1 A sin 1 B sin 1 C

.

2

2

2

≤ 18

Solution: Let u = sin 1 A sin 1 B sin 1 C. Apply formula (3.40) to the first two terms in u to get 2

2

2

u = − 1 (cos 1 ( A

( A

C

(cos 1 ( A

( A

( A

2

2

+ B) − cos 12

− B)) sin 12

= 12

2

− B) − cos 12

+ B)) cos 12

+ B) ,

since sin 1 C

( A

2

= cos 12

+ B), as we saw in Example 3.18. Multiply both sides by 2 to get

cos2 1 ( A

( A

( A

2

+ B) − cos 12

− B) cos 12

+ B) + 2 u = 0 ,

after rearranging the terms. Notice that the expression above is a quadratic equation in the term

cos 1 ( A

2

+ B). So by the quadratic formula,

cos 1 ( A

( A

2

− B) ±

cos2 12

− B) − 4(1)(2 u)

cos 1 ( A

,

2

+ B) =

2

which has a real solution only if the quantity inside the square root is nonnegative. But we know

that cos 1 ( A

2

+ B) is a real number (and, hence, a solution exists), so we must have

cos2 1 ( A

cos2 1 ( A

A sin 1 B sin 1 C

.

2

− B) − 8 u ≥ 0

⇒

u ≤ 18

2

− B) ≤ 18

⇒

sin 12

2

2

≤ 18

Example 3.20

For any triangle △ ABC, show that 1 < cos A + cos B + cos C ≤ 3 .

2

Solution: Since 0◦ < A, B, C < 180◦, the sines of 1 A, 1 B, and 1 C are all positive, so 2

2

2

cos A + cos B + cos C = 1 + 4 sin 1 A sin 1 B sin 1 C

2

2

2

> 1

by Example 3.18. Also, by Examples 3.18 and 3.19 we have cos A + cos B + cos C = 1 + 4 sin 1 A sin 1 B sin 1 C

.

2

2

2

≤ 1 + 4 · 18 = 32

Hence, 1 < cos A + cos B + cos C ≤ 3 .

2

Other Identities • Section 3.4

85

Example 3.21

Recall Snell’s law from Example 3.12 in Section 3.2: n 1 sin θ 1 = n 2 sin θ 2. Use it to show that the p-polarization transmission Fresnel coefficient defined by

2 n 1 cos θ 1

t 1 2 p =

(3.45)

n 2 cos θ 1 + n 1 cos θ 2

can be written as:

2 cos θ 1 sin θ 2

t 1 2 p =

.

sin ( θ 1 + θ 2) cos ( θ 1 − θ 2)

Solution: Multiply the top and bottom of t 1 2 p by sin θ 1 sin θ 2 to get:

2 n 1 cos θ 1

sin θ 1 sin θ 2

t 1 2 p =

·

n 2 cos θ 1 + n 1 cos θ 2

sin θ 1 sin θ 2

2 ( n 1 sin θ 1) cos θ 1 sin θ 2

= ( n 2 sin θ 2) sin θ 1 cos θ 1 + ( n 1 sin θ 1) sin θ 2 cos θ 2

2 cos θ 1 sin θ 2

=

(by Snell’s law)

sin θ 1 cos θ 1 + sin θ 2 cos θ 2

2 cos θ 1 sin θ 2

=

(by the double-angle formula)

1 (sin 2 θ

2

1 + sin 2 θ 2)

2 cos θ 1 sin θ 2

=

(by formula (3.41))

1 (2 sin 1 (2 θ

(2 θ

2

2

1 + 2 θ 2) cos 1

2

1 − 2 θ 2))

2 cos θ 1 sin θ 2

= sin ( θ 1 + θ 2) cos ( θ 1 − θ 2)

Example 3.22

In an AC electrical circuit, the instantaneous power p( t) delivered to the entire circuit in the sinu-

soidal steady state at time t is given by

p( t) = v( t) i( t) ,

where the voltage v( t) and current i( t) are given by

v( t) = Vm cos ωt ,

i( t) = Im cos ( ωt + φ) ,

for some constants Vm, Im, ω, and φ. Show that the instantaneous power can be written as

p( t) = 1 V

V

2

m I m cos φ + 1

2

m I m cos (2 ωt + φ) .

Solution: By definition of p( t), we have

p( t) = Vm Im cos ωt cos ( ωt + φ)

= Vm Im · 1 (cos (2 ωt

2

+ φ) + cos (− φ))

(by formula (3.43))

= 1 V

V

2

m I m cos φ + 1

2

m I m cos (2 ωt + φ)

(since cos (− φ) = cos φ) .

86

Chapter 3 • Identities

§3.4

Exercises

1. Prove formula (3.38).

2. Prove formula (3.39).

3. Prove formula (3.40).

4. Prove formula (3.41).

5. Prove formula (3.42).

6. Prove formula (3.44).

a + b

cos 1 ( A − B)

7. Prove Mollweide’s second equation: For any triangle △ ABC,

=

2

.

c

sin 1 C

2

8. Continuing Example 3.21, use Snell’s law to show that the p-polarization reflection Fresnel coefficient

n 2 cos θ 1 − n 1 cos θ 2

r 1 2 p =

(3.46)

n 2 cos θ 1 + n 1 cos θ 2

can be written as:

tan ( θ 1 − θ 2)

r 1 2 p = tan ( θ 1 + θ 2)

9. There is a more general form for the instantaneous power p( t) = v( t) i( t) in an electrical circuit than the one in Example 3.22. The voltage v( t) and current i( t) can be given by

v( t) = Vm cos ( ωt + θ) ,

i( t) = Im cos ( ωt + φ) ,

where θ is called the phase angle.4 Show that p( t) can be written as

p( t) = 1 V

V

2

m I m cos ( θ − φ) + 1

2

m I m cos (2 ωt + θ + φ) .

For Exercises 10-15, prove the given identity or inequality for any triangle △ ABC.

10. sin A + sin B + sin C = 4 cos 1 A cos 1 B cos 1 C

2

2

2

( Hint: Mimic Example 3.18 using (sin A + sin B) + (sin C − sin ( A + B + C)) . ) 11. cos A + cos ( B − C) = 2 sin B sin C

12. sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C

( Hints: Group sin 2 B and sin 2 C together, use the double-angle formula for sin 2 A, use Exercise

11. )

a − b

sin A − sin B

13.

=

a + b

sin A + sin B

s ( s − a)

( s − b) ( s − c)

14. cos 1 A

and sin 1 A

, where s

( a

2

=

=

= 1

+ b + c)

bc

2

bc

2

( Hint: Use the Law of Cosines to show that 2 bc (1 + cos A) = 4 s ( s − a) . )

15. 1 (sin A

( A

2

+ sin B) ≤ sin 12

+ B)

( Hint: Show that sin 1 ( A

(sin A

2

+ B) − 12

+ sin B) ≥ 0 . )

16. In Example 3.20, which angles A, B, C give the maximum value of cos A + cos B + cos C ?

4Though it does not matter for this exercise, none of the angles in these formulas are measured in degrees. We

will discuss their unit of measurement in Chapter 4.

4 Radian Measure

4.1 Radians and Degrees

So far we have been using degrees as our unit of measurement for angles. However, there is

another way of measuring angles that is often more convenient. The idea is simple: associate

a central angle of a circle with the arc that it intercepts.

Consider a circle of radius r > 0, as in Figure 4.1.1. In geometry you learned that the

circumference C of the circle is C = 2 π r, where π = 3.14159265....

AB = 1 C

r

AB

C

4

= π 2

= 12 = π r

AB = C = 2 π r

B

90◦

180◦

360◦

A

A

B

A

O

O

O

B

(a) θ = 90◦

(b) θ = 180◦

(c) θ = 360◦

Figure 4.1.1

Angle θ and intercepted arc AB on circle of circumference C = 2 πr

In Figure 4.1.1 we see that a central angle of 90◦ cuts off an arc of length π r, a central

2

angle of 180◦ cuts off an arc of length π r, and a central angle of 360◦ cuts off an arc of length

2 π r, which is the same as the circumference of the circle. So associating the central angle

with its intercepted arc, we could say, for example, that

360◦

“equals”

2 π r

(or 2 π ‘radiuses’).

The radius r was arbitrary, but the 2 π in front of it stays the same. So instead of using the

awkward “radiuses” or “radii”, we use the term radians:

360◦ = 2 π radians

(4.1)

The above relation gives us any easy way to convert between degrees and radians:

π

Degrees to radians:

x degrees

=

· x radians

(4.2)

180

180

Radians to degrees:

x radians

=

· x

degrees

(4.3)

π

87

88

Chapter 4 • Radian Measure

§4.1

Formula (4.2) follows by dividing both sides of equation (4.1) by 360, so that 1◦ = 2 π

360 = π

180

radians, then multiplying both sides by x. Formula (4.3) is similarly derived by dividing

both sides of equation (4.1) by 2 π then multiplying both sides by x.

The statement θ = 2 π radians is usually abbreviated as θ = 2 π rad, or just θ = 2 π when it is clear that we are using radians. When an angle is given as some multiple of π, you can

assume that the units being used are radians.

Example 4.1

Convert 18◦ to radians.

Solution: Using the conversion formula (4.2) for degrees to radians, we get

π

π

18◦ =

· 18 =

rad .

180

10

Example 4.2

Convert π radians to degrees.

9

Solution: Using the conversion formula (4.3) for radians to degrees, we get

π

180

π

rad =

·

= 20◦ .

9

π

9

Table 4.1

Commonly used angles in radians

Degrees

Radians

Degrees

Radians

Degrees

Radians

Degrees

Radians

π

3 π

0◦

0

90◦

180◦

π

270◦

2

2

π

2 π

7 π

5 π

30◦

120◦

210◦

300◦

6

3

6

3

π

3 π

5 π

7 π

45◦

135◦

225◦

315◦

4

4

4

4

π

5 π

4 π

11 π

60◦

150◦

240◦

330◦

3

6

3

6

Table 4.1 shows the conversion between degrees and radians for some

r

common angles. Using the conversion formula (4.3) for radians to degrees,

θ

we see that

180

O

r

1 radian =

degrees ≈ 57.3◦ .

π

Formally, a radian is defined as the central angle in a circle of radius

θ = 1 radian

r which intercepts an arc of length r, as in Figure 4.1.2. This definition

Figure 4.1.2

does not depend on the choice of r (imagine resizing Figure 4.1.2).

Radians and Degrees • Section 4.1

89

One reason why radians are used is that the scale is smaller than for degrees. One revolu-

tion in radians is 2 π ≈ 6.283185307, which is much smaller than 360, the number of degrees

in one revolution. The smaller scale makes the graphs of trigonometric functions (which we

will discuss in Chapter 5) have similar scales for the horizontal and vertical axes. Another

reason is that often in physical applications the variables being used are in terms of arc

length, which makes radians a natural choice.

The default mode in most scientific calculators is to use degrees for entering angles. On

✄

many calculators there is a button labeled DRG for switching between degree mode (D),

✂

✁

radian mode (R), and gradian mode (G).1 On some graphing calculators, such as the the TI-

✄

83, there is a MODE button for changing between degrees and radians. Make sure that your

✂

✁

calculator is in the correct angle mode before entering angles, or your answers will likely be

way off. For example,

sin 4◦ =

0.0698 ,

sin (4 rad) = −0.7568 ,

so the values are not only off in magnitude, but do not even have the same sign. Using your

✄

✄

✄

calculator’s sin−1 , cos−1 , and tan−1 buttons in radian mode will of course give you the angle

✂

✁ ✂

✁

✂

✁

as a decimal, not an expression in terms of π.

You should also be aware that the math functions in many computer programming lan-

guages use radians, so you would have to write your own angle conversions.2

Exercises

For Exercises 1-5, convert the given angle to radians.

1. 4◦

2. 15◦

3. 130◦

4. 275◦

5. −108◦

For Exercises 6-10, convert the given angle to degrees.

π

11 π

29 π

6. 4 rad

7.

rad

8.

rad

9.

rad

10. 35 rad

5

9

30

11. Put your calculator in radian mode and take the cosine of 0. Whatever the answer is, take its

cosine. Then take the cosine of the new answer. Keep repeating this. On most calculators after

about 50-60 iterations you should start to see the same answer repeating. What is that number?

Try starting with a number different from 0. Do you get the same answer repeating after roughly

the same number of iterations as before? Try the same procedure in degree mode, starting with 0◦.

Does the same thing happen? If so, does it take fewer iterations for the answer to start repeating

than in radian mode, or more?

1A gradian is defined as 1 of a circle, i.e. there are 400 gradians in one revolution. Compared to the more

400

common 360◦ in one revolution, gradians appear to be easier to work with, since a right angle is 100 gradians

(thus making integer multiples of a right angle easier to remember). Outside of a few specialized areas (e.g.

artillery calculations), gradians are nevertheless not widely used today.

2One exception is Octave, which has functions cosd(), sind(), tand() that take angles in degrees as parameters,

in addition to the usual cos(), sin(), tan() functions which use radians.

90

Chapter 4 • Radian Measure

§4.2

4.2 Arc Length

In Section 4.1 we saw that one revolution has a radian measure of 2 π rad. Note that 2 π is

the ratio of the circumference (i.e. total arc length) C of a circle to its radius r:

2 π r

C

total arc length

Radian measure of 1 revolution = 2 π =

=

=

r

r

radius

Clearly, that ratio is independent of r. In general, the radian measure of an angle is the

ratio of the arc length cut off by the corresponding central angle in a circle to the radius of

the circle, independent of the radius.

To see this, recall our formal definition of a radian: the central angle in a circle of radius r

which intercepts an arc of length r. So suppose that we have a circle of radius r and we place

a central angle with radian measure 1 on top of another central angle with radian measure

1, as in Figure 4.2.1(a). Clearly, the combined central angle of the two angles has radian

measure 1 + 1 = 2, and the combined arc length is r + r = 2 r.

r

r/2

2

r

1

r/2

1 1

1/2

r

r

(a) 2 radians

(b) 1 radian

2

Figure 4.2.1

Radian measure and arc length

Now suppose that we cut the angle with radian measure 1 in half, as in Figure 4.2.1(b).

Clearly, this cuts the arc length r in half as well. Thus, we see that

Angle = 1 radian

⇒

arc length = r ,

Angle = 2 radians ⇒

arc length = 2 r ,

Angle = 1 radian

r ,

2

⇒

arc length = 12

and in general, for any θ ≥ 0,

Angle = θ radians ⇒

arc length = θ r ,

so that

arc length

θ =

.

radius

Arc Length • Section 4.2

91

Intuitively, it is obvious that shrinking or magnifying a circle preserves the measure of a

central angle even as the radius changes. The above discussion says more, namely that the

ratio of the length s of an intercepted arc to the radius r is preserved, precisely because that

ratio is the measure of the central angle in radians (see Figure 4.2.2).

s = r′ θ

s = rθ

θ

θ

O

r′

O

r

(a) Angle θ, radius r

(b) Angle θ, radius r′

Figure 4.2.2

Circles with the same central angle, different radii

We thus get a simple formula for the length of an arc:

In a circle of radius r, let s be the length of an arc intercepted by a central angle with

radian measure θ ≥ 0. Then the arc length s is:

s = r θ

(4.4)

Example 4.3

In a circle of radius r = 2 cm, what is the length s of the arc intercepted by a central angle of measure

θ = 1.2 rad ?

Solution: Using formula (4.4), we get:

s = r θ = (2)(1.2) = 2.4 cm

Example 4.4

In a circle of radius r = 10 ft, what is the length s of the arc intercepted by a central angle of measure

θ = 41◦ ?

Solution: Using formula (4.4) blindly with θ = 41◦, we would get s = r θ = (10)(41) = 410 ft. But this impossible, since a circle of radius 10 ft has a circumference of only 2 π (10) ≈ 62.83 ft! Our error was

in using the angle θ measured in degrees, not radians. So first convert θ = 41◦ to radians, then use

s = r θ:

π

θ = 41◦ =

· 41 = 0.716 rad

⇒

s = r θ = (10)(0.716) = 7.16 ft

180

Note that since the arc length s and radius r are usually given in the same units, radian

measure is really unitless, since you can think of the units canceling in the ratio s , which is

r

just θ. This is another reason why radians are so widely used.

92

Chapter 4 • Radian Measure

§4.2

Example 4.5

A central angle in a circle of radius 5 m cuts off an arc of length 2 m. What is the measure of the

angle in radians? What is the measure in degrees?

Solution: Letting r = 5 and s = 2 in formula (4.4), we get:

s

2

θ =

=

= 0.4 rad

r

5

In degrees, the angle is:

180

θ = 0.4 rad =

· 0.4 = 22.92◦

π

For central angles θ > 2 π rad, i.e. θ > 360◦, it may not be clear what is meant by the inter-

cepted arc, since the angle is larger than one revolution and hence “wraps around” the circle

more than once. We will take the approach that such an arc consists of the full circumference

plus any additional arc length determined by the angle. In other words, formula (4.4) is still

valid for angles θ > 2 π rad.

What about negative angles? In this case using s = r θ would mean that the arc length is

negative, which violates the usual concept of length. So we will adopt the convention of only

using nonnegative central angles when discussing arc length.

Example 4.6

A rope is fastened to a wall in two places 8 ft apart at the same height. A

A

cylindrical container with a radius of 2 ft is pushed away from the wall as

far as it can go while being held in by the rope, as in Figure 4.2.3 which

B

shows the top view. If the center of the container is 3 feet away from the

4

point on the wall midway between the ends of the rope, what is the length

2

L of the rope?

θ

D

C

3

E

Solution: We see that, by symmetry, the total length of the rope is L =

2 ( AB + BC). Also, notice that △ ADE is a right triangle, so the hypotenuse

4

has length AE = DE 2 + D A 2 = 32 + 42 = 5 ft, by the Pythagorean The-

orem. Now since AB is tangent to the circular container, we know that

∠ ABE is a right angle. So by the Pythagorean Theorem we have

Figure 4.2.3

AB =

AE 2 − BE 2 =

52 − 22 =

21 ft.

By formula (4.4) the arc BC has length BE · θ, where θ = ∠ BEC is the

supplement of ∠ AED + ∠ AEB. So since

4

BE

2

tan ∠ AED =

⇒ ∠ AED = 53.1◦ and

cos ∠ AEB =

=

⇒ ∠ AEB = 66.4◦ ,

3

AE

5

we have

θ = ∠ BEC = 180◦ − (∠ AED + ∠ AEB) = 180◦ − (53.1◦ + 66.4◦) = 60.5◦ .

Converting to radians, we get θ = π

180 · 60.5 = 1.06 rad. Thus,

L = 2( AB + · BC) = 2( 21 + BE · θ) = 2( 21 + (2)(1.06)) = 13.4 ft .

Arc Length • Section 4.2

93

Example 4.7

The centers of two belt pulleys, with radii of 5 cm and 8 cm, respectively, are 15 cm apart. Find the

total length L of the belt around the pulleys.

Solution: In Figure 4.2.4 we see that, by symmetry, L = 2 ( DE + EF + FG).

F

E

5

C

5

3

D

G

A

B

15

Figure 4.2.4

Belt pulleys with radii 5 cm and 8 cm

First, at the center B of the pulley with radius 8, draw a circle of radius 3, which is the difference

in the radii of the two pulleys. Let C be the point where this circle intersects BF. Then we know that

the tangent line AC to this smaller circle is perpendicular to the line segment BF. Thus, ∠ ACB is a

right angle, and so the length of AC is

AC =

AB 2 − BC 2 =

152 − 32 =

216 = 6 6

by the Pythagorean Theorem. Now since AE ⊥ EF and EF ⊥ CF and CF ⊥ AC, the quadrilateral

AEFC must be a rectangle. In particular, EF = AC, so EF = 6 6.

By formula (4.4) we know that DE = E A · ∠ D AE and FG = BF · ∠ GBF, where the angles are measured in radians. So thinking of angles in radians (using π rad = 180◦), we see from Figure 4.2.4

that

π

π

∠ D AE = π − ∠ E AC − ∠ BAC = π −

− ∠ BAC =

− ∠ BAC ,

2

2

where

BC

3

sin ∠ BAC =

=

= 0.2

⇒

∠ BAC = 0.201 rad.

AB

15

Thus, ∠ D AE = π 2 − 0.201 = 1.37 rad. So since AE and BF are parallel, we have ∠ ABC = ∠ D AE =

1.37 rad. Thus, ∠ GBF = π − ∠ ABC = π − 1.37 = 1.77 rad. Hence,

L = 2 ( DE + EF + FG) = 2 (5 (1.37) + 6 6 + 8 (1.77)) = 71.41 cm .

94

Chapter 4 • Radian Measure

§4.2

Exercises

For Exercises 1-4, find the length of the arc cut off by the given central angle θ in a circle of radius r.

1. θ = 0.8 rad, r = 12 cm

2. θ = 171◦, r = 8 m

3. θ = π rad, r = 11 in

4. A central angle in a circle of radius 2 cm cuts off an arc of length 4.6 cm. What is the measure of

the angle in radians? What is the measure of the angle in degrees?

5. The centers of two belt pulleys, with radii of 3 inches and 6 inches, respectively, are 13 inches

apart. Find the total length L of the belt around the pulleys.

6. In Figure 4.2.5 one end of a 4 ft iron rod is attached to the center of a pulley with radius 0.5 ft.

The other end is attached at a 40◦ angle to a wall, at a spot 6 ft above the lower end of a steel wire

supporting a box. The other end of the wire comes out of the wall straight across from the top of

the pulley. Find the length L of the wire from the wall to the box.

4

4

40◦

2

40◦

6

6

Figure 4.2.5

Exercise 6

Figure 4.2.6

Exercise 7

7. Figure 4.2.6 shows the same setup as in Exercise 6 but now the wire comes out of the wall 2 ft above where the rod is attached. Find the length L of the wire from the wall to the box.

8. Find the total length L of the figure eight shape

2

2

in Figure 4.2.7.

9. Repeat Exercise 8 but with the circle at A hav-

A

B

10

ing a radius of 3 instead of 2. ( Hint: Draw a

circle of radius 5 centered at A, then draw a

tangent line to that circle from B. )

Figure 4.2.7

10. Suppose that in Figure 4.2.7 the lines do not criss-cross but instead go straight across, as in a

belt pulley system. Find the total length L of the resulting shape.

11. Find the lengths of the two arcs cut off by a chord of length 3 in a circle of radius 2.

12. Find the perimeter of a regular dodecagon (i.e. a 12-sided polygon with sides of equal length)

inscribed inside a circle of radius 1 . Compare it to the circumference of the circle.

2

Area of a Sector • Section 4.3

95

4.3 Area of a Sector

In geometry you learned that the area of a circle of radius r is πr 2. We

r

will now learn how to find the area of a sector of a circle. A sector is

θ

the region bounded by a central angle and its intercepted arc, such as the

shaded region in Figure 4.3.1.

Let θ be a central angle in a circle of radius r and let A be the area of its

sector. Similar to arc length, the ratio of A to the area of the entire circle

Figure 4.3.1

is the same as the ratio of θ to one revolution. In other words, again using

radian measure,

area of sector

sector angle

A

θ

=

⇒

=

.

area of entire circle

one revolution

π r 2

2 π

Solving for A in the above equation, we get the following formula:

In a circle of radius r, the area A of the sector inside a central angle θ is

A = 1 r 2 θ ,

(4.5)

2

where θ is measured in radians.

Example 4.8

Find the area of a sector whose angle is π rad in a circle of radius 4 cm.

5

Solution: Using θ = π and r

5

= 4 in formula (4.5), the area A of the sector is

A = 1 r 2 θ

(4)2

cm2 .

2

= 12

· π 5 = 8 π

5

Example 4.9

Find the area of a sector whose angle is 117◦ in a circle of radius 3.5 m.

Solution: As with arc length, we have to make sure that the angle is measured in radians or else

the answer will be way off. So converting θ = 117◦ to radians and using r = 3.5 in formula (4.5) for the area A of the sector, we get

π

θ = 117◦ =

· 117 = 2.042 rad

⇒

A = 1 r 2 θ = 1 (3.5)2 (2.042) = 12.51 m2 .

180

2

2

For a sector whose angle is θ in a circle of radius r, the length of the arc cut off by that

angle is s = r θ. Thus, by formula (4.5) the area A of the sector can be written as:3

A = 1 rs

(4.6)

2

Note: The central angle θ that intercepts an arc is sometimes called the angle subtended

by the arc.

3In some texts this formula is taken as a result from elementary geometry and then used to prove formula

(4.5).

96

Chapter 4 • Radian Measure

§4.3

Example 4.10

Find the area of a sector whose arc is 6 cm in a circle of radius 9 cm.

Solution: Using s = 6 and r = 9 in formula (4.6) for the area A, we get

A = 1 rs

(9) (6)

2

= = 12

= 27 cm2 .

Note that the angle subtended by the arc is θ = s

rad.

r = 2

3

Example 4.11

Find the area K inside the belt pulley system from Example 4.7 in Section 4.2.

Solution: Recall that the belt pulleys have radii of 5 cm and 8 cm, and their centers are 15 cm apart.

We showed in Example 4.7 that EF = AC = 6 6, ∠ D AE = 1.37 rad, and ∠ GBF = 1.77 rad. We see from Figure 4.3.2 that, by symmetry, the total area K enclosed by the belt is twice the area above the line DG, that is,

K = 2((Area of sector D AE) + (Area of rectangle AEFC)

+ (Area of triangle △ ABC) + (Area of sector GBF)) .

F

6 6

E

5

C

5

6 6

1.77

3

1.37

D

G

A

B

15

Figure 4.3.2

Belt pulleys with radii 5 cm and 8 cm

Since AEFC is a rectangle with sides 5 and 6 6, its area is 30 6. And since △ ABC is a right

triangle whose legs have lengths 3 and 6 6, its area is 1 (3) (6 6)

2

= 9 6. Thus, using formula (4.5)

for the areas of sectors D AE and GBF, we have

K = 2 (Area of sector D AE) + 30 6 + 9 6 + (Area of sector GBF)

= 2 1 (5)2 (1.37)

(8)2 (1.77)

2

+ 30 6 + 9 6 + 12

= 338.59 cm2 .

Area of a Sector • Section 4.3

97

So far we have dealt with the area cut off by a central angle. How

a

would you find the area of a region cut off by an inscribed angle, such

r

as the shaded region in Figure 4.3.3? In this picture, the center of the

circle is inside the inscribed angle, and the lengths a and b of the two

chords are given, as is the radius r of the circle. Drawing line segments

b

from the center of the circle to the endpoints of the chords indicates

how to solve this problem: add up the areas of the two triangles and

Figure 4.3.3

the sector formed by the central angle. The areas and angles of the two

triangles can be determined (since all three sides are known) using methods from Chapter

2. Also, recall (Theorem 2.4 in Section 2.5) that a central angle has twice the measure of any

inscribed angle which intercepts the same arc. In the exercises you will be asked to solve

problems like this (including the cases where the center of the circle is outside or on the

inscribed angle).

Another type of region we can consider is a segment of a circle,

A

which is the region between a chord and the arc it cuts off. In Figure

r

4.3.4 the segment formed by the chord AB is the shaded region

between the arc AB and the triangle

θ

△ OAB. By formula (2.23)

B

r

in Section 2.4 for the area of a triangle given two sides and their

O

included angle, we know that

area of △ OAB = 1 ( r)( r) sin θ

r 2 sin θ .

2

= 12

Figure 4.3.4

Thus, since the area K of the segment is the area of the sector

AOB minus the area of the triangle △ OAB, we have

area K of segment AB = 1 r 2 θ

r 2 sin θ

r 2 ( θ

2

− 12

= 12

− sin θ) .

(4.7)

Note that as a consequence of formula (4.7) we must have θ > sin θ for 0 < θ ≤ π (measured in radians), since the area of a segment is positive for those angles.

Example 4.12

Find the area of the segment formed by a chord of length 3 in a circle of

radius 2.

3

2

Solution: Figure 4.3.5 shows the segment formed by a chord of length 3 in

θ

a circle of radius r = 2. We can use the Law of Cosines to find the subtended

2

central angle θ:

22 + 22 − 32

cos θ =

= −0.125

⇒

θ = 1.696 rad

2 (2) (2)

Figure 4.3.5

Thus, by formula (4.7) the area K of the segment is:

K = 1 r 2 ( θ

(2)2 (1.696

2

− sin θ) = 12

− sin 1.696) = 1.408

98

Chapter 4 • Radian Measure

§4.3

Example 4.13

The centers of two circles are 7 cm apart, with one circle having a radius of 5 cm and the other a

radius of 3 cm. Find the area K of their intersection.

Solution: In Figure 4.3.6(a), we see that the intersection of the two circles is the union of the seg-

ments formed by the chord CD in each circle. Thus, once we determine the angles ∠ C AD and ∠ CBD

we can calculate the area of each segment and add those areas together to get K .

C

5

4

A

B

7

C

5

4

D

α

β

A

B

7

(a) ∠ BAC = 1

(b) Triangle △ ABC

2 ∠ C AD, ∠ ABC = 1

2 ∠ CBD

Figure 4.3.6

By symmetry, we see that ∠ BAC = 12 ∠ CAD and ∠ ABC = 12 ∠ CBD. So let α = ∠ BAC and β =

∠ ABC, as in Figure 4.3.6(b). By the Law of Cosines, we have

72 + 52 − 42

cos α =

= 0.8286

⇒

α = 0.594 rad

⇒

∠ C AD = 2(0.594) = 1.188 rad

2 (7) (5)

72 + 42 − 52

cos β =

= 0.7143

⇒

β = 0.775 rad

⇒

∠ CBD = 2(0.775) = 1.550 rad

2 (7) (4)

Thus, the area K is

K = (Area of segment CD in circle at A) + (Area of segment CD in circle at B)

= 1 (5)2 (1.188

(4)2 (1.550

2

− sin 1.188) + 12

− sin 1.550)

= 7.656 cm2 .

Exercises

For Exercises 1-3, find the area of the sector for the given angle θ and radius r.

1. θ = 2.1 rad, r = 1.2 cm

2. θ = 3 π rad, r

7

= 3.5 ft

3. θ = 78◦, r = 6 m

4. The centers of two belt pulleys, with radii of 3 cm and 6 cm, respectively, are 13 cm apart. Find

the total area K enclosed by the belt.

5. In Exercise 4 suppose that both belt pulleys have the same radius of 6 cm. Find the total area K

enclosed by the belt.

Area of a Sector • Section 4.3

99

6. Find the area enclosed by the figure eight in Exercise 8 from Section 4.2.

For Exercises 7-9, find the area of the sector for the given radius r and arc length s.

7. r = 5 cm, s = 2 cm

8. r = a, s = a

9. r = 1 cm, s = π cm

For Exercises 10-12, find the area of the segment formed by a chord of length a in a circle of radius r.

10. a = 4 cm, r = 4 cm

11. a = 1 cm, r = 5 cm

12. a = 2 cm, r = 5 cm

13. Find the area of the shaded region in Figure 4.3.7.

4

7

8

3

9

5

6

6

5

Figure 4.3.7 Exercise 13

Figure 4.3.8 Exercise 14

Figure 4.3.9 Exercise 15

14. Find the area of the shaded region in Figure 4.3.8. ( Hint: Draw two central angles. )

15. Find the area of the shaded region in Figure 4.3.9.

16. The centers of two circles are 4 cm apart, with one circle having a radius of 3 cm and the other a

radius of 2 cm. Find the area of their intersection.

17. Three circles with radii of 4 m, 2 m, and 1 m are externally tangent to each other. Find the area

of the curved region between the circles, as in Figure 4.3.10. ( Hint: Connect the centers of the

circles. )

1

r

4

2

r

r

Figure 4.3.10

Exercise 17

Figure 4.3.11

Exercise 18

18. Show that the total area enclosed by the loop around the three circles of radius r in Figure 4.3.11

is ( π + 6 + 3) r 2.

19. For a fixed central angle θ, how much does the area of its sector increase when the radius of the

circle is doubled? How much does the length of its intercepted arc increase?

100

Chapter 4 • Radian Measure

§4.4

4.4 Circular Motion: Linear and Angular Speed

Radian measure and arc length can be applied to the study

time t > 0

of circular motion. In physics the average speed of an object

is defined as:

distance s

θ

distance traveled

average speed =

r

time elapsed

time t = 0

So suppose that an object moves along a circle of radius r,

traveling a distance s over a period of time t, as in Figure

Figure 4.4.1

4.4.1. Then it makes sense to define the (average) linear

speed ν of the object as:

s

ν =

(4.8)

t

Let θ be the angle swept out by the object in that period of time. Then we define the

(average) angular speed ω of the object as:

θ

ω =

(4.9)

t

Angular speed gives the rate at which the central angle swept out by the object changes as

the object moves around the circle, and it is thus measured in radians per unit time. Linear

speed is measured in distance units per unit time (e.g. feet per second). The word linear is

used because straightening out the arc traveled by the object along the circle results in a line

of the same length, so that the usual definition of speed as distance over time can be used.

We will usually omit the word average when discussing linear and angular speed here.4

Since the length s of the arc cut off by a central angle θ in a circle of radius r is s = r θ, we see that

s

r θ

θ

ν =

=

=

· r ,

t

t

t

so that we get the following relation between linear and angular speed:

ν = ω r

(4.10)

4Many trigonometry texts assume uniform motion, i.e. constant speeds. We do not make that assumption. Also,

many texts use the word velocity instead of speed. Technically they are not the same; velocity has a direction

and a magnitude, whereas speed is just a magnitude.

Circular Motion: Linear and Angular Speed • Section 4.4

101

Example 4.14

An object sweeps out a central angle of π radians in 0.5 seconds as it moves along a circle of radius 3

3

m. Find its linear and angular speed over that time period.

Solution: Here we have t = 0.5 sec, r = 3 m, and θ = π rad. So the angular speed ω is

3

π rad

θ

3

2 π

ω =

=

⇒

ω =

rad/sec ,

t

0.5 sec

3

and thus the linear speed ν is

2 π

ν = ω r =

rad/sec (3 m)

⇒

ν = 2 π m/sec .

3

Note that the units for ω are rad/sec and the units of ν are m/sec. Recall that radians are actually

unitless, which is why in the formula ν = ω r the radian units disappear.

Example 4.15

An object travels a distance of 35 ft in 2.7 seconds as it moves along a circle of radius 2 ft. Find its

linear and angular speed over that time period.

Solution: Here we have t = 2.7 sec, r = 2 ft, and s = 35 ft. So the linear speed ν is

s

35 feet

ν =

=

⇒

ν = 12.96 ft/sec ,

t

2.7 sec

and thus the angular speed ω is given by

ν = ω r

⇒

12.96 ft/sec = ω(2 ft)

⇒

ω = 6.48 rad/sec .

Example 4.16

An object moves at a constant linear speed of 10 m/sec around a circle of radius 4 m. How large of a

central angle does it sweep out in 3.1 seconds?

Solution: Here we have t = 3.1 sec, ν = 10 m/sec, and r = 4 m. Thus, the angle θ is given by

s

ν t

(10 m/sec) (3.1 sec)

s = r θ

⇒

θ =

=

=

= 7.75 rad .

r

r

4 m

In many physical applications angular speed is given in revolutions per minute, abbrevi-

ated as rpm. To convert from rpm to, say, radians per second, notice that since there are 2 π

radians in one revolution and 60 seconds in one minute, we can convert N rpm to radians

per second by “canceling the units” as follows:

✘✘

rev

2 π rad

1 ✟✟

min

N · 2 π

N rpm = N

·

·

=

rad/sec

✟✟

min

1 ✘✘

rev

60 sec

60

This works because all we did was multiply by 1 twice. Converting to other units for angular

speed works in a similar way. Going in the opposite direction, say, from rad/sec to rpm, gives:

N · 60

N rad/sec =

rpm

2 π

102

Chapter 4 • Radian Measure

§4.4

Example 4.17

A gear with an outer radius of r 1 = 5 cm moves in the clockwise direction, causing an interlocking

gear with an outer radius of r 2 = 4 cm to move in the counterclockwise direction at an angular speed

of ω 2 = 25 rpm. What is the angular speed ω 1 of the larger gear?

Solution: Imagine a particle on the outer radius of each gear.

r 1 = 5 cm

After the gears have rotated for a period of time t > 0, the cir-

cular displacement of each particle will be the same. In other

ω 2 = 25 rpm

r 2

words, s

= 4 cm

1 = s 2, where s 1 and s 2 are the distances traveled by

the particles on the gears with radii r 1 and r 2, respectively.

But s 1 = ν 1 t and s 2 = ν 2 t, where ν 1 and ν 2 are the linear

speeds of the gears with radii r 1 and r 2, respectively. Thus,

ν 1 t = ν 2 t

⇒

ν 1 = ν 2 ,

so by formula (4.10) we get the fundamental relation between

the two gears:

Figure 4.4.2

ω 1 r 1 = ω 2 r 2

(4.11)

Note that this holds for any two gears. So in our case, we have

ω 1 (5) = (25)(4)

⇒

ω 1 = 20 rpm .

Exercises

For Exercises 1-6, assume that a particle moves along a circle of radius r for a period of time t. Given

either the arc length s or the central angle θ swept out by the particle, find the linear and angular

speed of the particle.

1. r = 4 m, t = 2 sec, θ = 3 rad

2. r = 8 m, t = 2 sec, θ = 3 rad

3. r = 7 m, t = 3.2 sec, θ = 172◦

4. r = 1 m, t = 1.6 sec, s = 3 m

5. r = 2 m, t = 1.6 sec, s = 6 m

6. r = 1.5 ft, t = 0.3 sec, s = 4 in

7. An object moves at a constant linear speed of 6 m/sec around a circle of radius 3.2 m. How large

of a central angle does it sweep out in 1.8 seconds?

8. Two interlocking gears have outer radii of 6 cm and 9 cm, respectively. If the smaller gear rotates

at 40 rpm, how fast does the larger gear rotate?

9. Three interlocking gears have outer radii of 2 cm, 3 cm, and 4 cm, respectively. If the largest gear

rotates at 16 rpm, how fast do the other gears rotate?

10. In Example 4.17, does equation (4.11) still hold if the radii r 1 and r 2 are replaced by the number of teeth N 1 and N 2, respectively, of the two gears as shown in Figure 4.4.2?

11. A 78 rpm music record has a diameter of 10 inches. What is the linear speed of a speck of dust

on the outer edge of the record in inches per second?

12. The centripetal acceleration α of an object moving along a circle of radius r with a linear speed ν

is defined as α = ν 2 . Show that α

r

= ω 2 r, where ω is the angular speed.

5 Graphing and Inverse Functions

The trigonometric functions can be graphed just like any other function, as we will now

show. In the graphs we will always use radians for the angle measure.

5.1 Graphing the Trigonometric Functions

The first function we will graph is the sine func-

y

tion. We will describe a geometrical way to create the

x 2 + y 2 = 1

( x, y) = (cos θ,sin θ)

graph, using the unit circle. This is the circle of radius

1

1 in the x y-plane consisting of all points ( x, y) which

s = r θ = θ

satisfy the equation x 2 + y 2 = 1.

θ

x

We see in Figure 5.1.1 that any point on the unit

0

1

circle has coordinates ( x, y) = (cos θ,sin θ), where θ

is the angle that the line segment from the origin to

( x, y) makes with the positive x-axis (by definition of

sine and cosine). So as the point ( x, y) goes around the

Figure 5.1.1

circle, its y-coordinate is sin θ.

We thus get a correspondence between the y-coordinates of points on the unit circle and

the values f ( θ) = sin θ, as shown by the horizontal lines from the unit circle to the graph of

f ( θ) = sin θ in Figure 5.1.2 for the angles θ = 0, π , π , π .

6

3

2

f ( θ)

1

1

π

2

π

f ( θ) = sin θ

3

π

6

θ

0

θ

1

0

π

π

π

2 π

5 π

π

x 2

6

3

2

+ y 2 = 1

3

6

Figure 5.1.2

Graph of sine function based on y-coordinate of points on unit circle

We can extend the above picture to include angles from 0 to 2 π radians, as in Figure 5.1.3.

This illustrates what is sometimes called the unit circle definition of the sine function.

103

104

Chapter 5 • Graphing and Inverse Functions

§5.1

y

f ( θ)

f ( θ) = sin θ

1

θ

θ

x

1

0

π

π

π

2 π

5 π

π

5 π

3 π

7 π

2 π

6

3

2

3

6

4

2

4

−1

x 2 + y 2 = 1

Figure 5.1.3

Unit circle definition of the sine function

Since the trigonometric functions repeat every 2 π radians (360◦), we get, for example, the

following graph of the function y = sin x for x in the interval [−2 π,2 π]:

y

y = sin x

1

x

0

π

π

−2 π

− π

3 π

π

5 π

3 π

7 π

2 π

− 7 π

4

− 3 π

2

− 5 π

4

− 3 π − π 2

− π 4

4

2

4

4

4

2

4

−1

Figure 5.1.4

Graph of y = sin x

To graph the cosine function, we could again use the unit circle idea (using the x-coordinate

of a point that moves around the circle), but there is an easier way. Recall from Section 1.5

that cos x = sin ( x + 90◦) for all x. So cos 0◦ has the same value as sin 90◦, cos 90◦ has the

same value as sin 180◦, cos 180◦ has the same value as sin 270◦, and so on. In other words,

the graph of the cosine function is just the graph of the sine function shifted to the left by

90◦ = π/2 radians, as in Figure 5.1.5:

y

y = cos x

1

x

0

π

π

−2 π

− π

3 π

π

5 π

3 π

7 π

2 π

− 7 π

4

− 3 π

2

− 5 π

4

− 3 π − π 2

− π 4

4

2

4

4

4

2

4

−1

Figure 5.1.5

Graph of y = cos x

To graph the tangent function, use tan x = sin x to get the following graph:

cos x

Graphing the Trigonometric Functions • Section 5.1

105

y

8

6

4

2

y = tan x

x

−2 π

− π

0

π

π

3 π

π

5 π

3 π

7 π

2 π

− 7 π

4

− 3 π

2

− 5 π

4

− 3 π − π 2

− π 4

4

2

4

4

4

2

4

−2

−4

−6

−8

Figure 5.1.6

Graph of y = tan x

Recall that the tangent is positive for angles in QI and QIII, and is negative in QII and

QIV, and that is indeed what the graph in Figure 5.1.6 shows. We know that tan x is not

defined when cos x = 0, i.e. at odd multiples of π : x

,

,

, etc. We can figure out

2

= ± π 2 ± 3 π

2

± 5 π

2

what happens near those angles by looking at the sine and cosine functions. For example,

for x in QI near π , sin x and cos x are both positive, with sin x very close to 1 and cos x very 2

close to 0, so the quotient tan x = sin x is a positive number that is very large. And the closer

cos x

x gets to π , the larger tan x gets. Thus, x

is a vertical asymptote of the graph of y

2

= π 2

= tan x.

Likewise, for x in QII very close to π , sin x is very close to 1 and cos x is negative and very

2

close to 0, so the quotient tan x = sin x is a negative number that is very large, and it gets

cos x

larger in the negative direction the closer x gets to π . The graph shows this. Similarly, we

2

get vertical asymptotes at x = − π , x

, and x

, as in Figure 5.1.6. Notice that the

2

= 3 π

2

= − 3 π

2

graph of the tangent function repeats every π radians, i.e. two times faster than the graphs

of sine and cosine repeat.

The graphs of the remaining trigonometric functions can be determined by looking at the

graphs of their reciprocal functions. For example, using csc x = 1 we can just look at the

sin x

graph of y = sin x and invert the values. We will get vertical asymptotes when sin x = 0,

namely at multiples of π: x = 0, ± π, ±2 π, etc. Figure 5.1.7 shows the graph of y = csc x, with the graph of y = sin x (the dashed curve) for reference.

106

Chapter 5 • Graphing and Inverse Functions

§5.1

y

4

y = csc x

3

2

1

x

0

π

π

−2 π

− π

3 π

π

5 π

3 π

7 π

2 π

− 7 π

4

− 3 π

2

− 5 π

4

− 3 π − π 2

− π 4

4

2

4

4

4

2

4

−1

−2

−3

−4

Figure 5.1.7

Graph of y = csc x

Likewise, Figure 5.1.8 shows the graph of y = sec x, with the graph of y = cos x (the dashed curve) for reference. Note the vertical asymptotes at x = ± π ,

. Notice also that the graph

2 ± 3 π

2

is just the graph of the cosecant function shifted to the left by π radians.

2

y

4

y = sec x

3

2

1

x

0

π

π

−2 π

− π

3 π

π

5 π

3 π

7 π

2 π

− 7 π

4

− 3 π

2

− 5 π

4

− 3 π − π 2

− π 4

4

2

4

4

4

2

4

−1

−2

−3

−4

Figure 5.1.8

Graph of y = sec x

Graphing the Trigonometric Functions • Section 5.1

107

The graph of y = cot x can also be determined by using cot x = 1 . Alternatively, we can

cot x

use the relation cot x = −tan ( x + 90◦) from Section 1.5, so that the graph of the cotangent

function is just the graph of the tangent function shifted to the left by π radians and then

2

reflected about the x-axis, as in Figure 5.1.9:

y

8

6

4

2

y = cot x

x

−2 π

− π

0

π

π

3 π

π

5 π

3 π

7 π

2 π

− 7 π

4

− 3 π

2

− 5 π

4

− 3 π − π 2

− π 4

4

2

4

4

4

2

4

−2

−4

−6

−8

Figure 5.1.9

Graph of y = cot x

Example 5.1

Draw the graph of y = −sin x for 0 ≤ x ≤ 2 π.

Solution: Multiplying a function by −1 just reflects its graph around the x-axis. So reflecting the

graph of y = sin x around the x-axis gives us the graph of y = −sin x:

y

y = −sin x

1

0

x

π

π

3 π

π

5 π

3 π

7 π

2 π

4

2

4

4

2

4

−1

Note that this graph is the same as the graphs of y = sin ( x ± π) and y = cos ( x + π ).

2

108

Chapter 5 • Graphing and Inverse Functions

§5.1

It is worthwhile to remember the general shapes of the graphs of the six trigonometric

functions, especially for sine, cosine, and tangent. In particular, the graphs of the sine and

cosine functions are called sinusoidal curves. Many phenomena in nature exhibit sinusoidal

behavior, so recognizing the general shape is important.

Example 5.2

Draw the graph of y = 1 + cos x for 0 ≤ x ≤ 2 π.

Solution: Adding a constant to a function just moves its graph up or down by that amount, depend-

ing on whether the constant is positive or negative, respectively. So adding 1 to cos x moves the graph

of y = cos x upward by 1, giving us the graph of y = 1 + cos x:

y

2

y = 1 + cos x

1

x

0

π

π

3 π

π

5 π

3 π

7 π

2 π

4

2

4

4

2

4

Exercises

For Exercises 1-12, draw the graph of the given function for 0 ≤ x ≤ 2 π.

1. y = −cos x

2. y = 1 + sin x

3. y = 2 − cos x

4. y = 2 − sin x

5. y = −tan x

6. y = −cot x

7. y = 1 + sec x

8. y = −1 − csc x

9. y = 2sin x

10. y = −3cos x

11. y = −2tan x

12. y = −2sec x

13. We can extend the unit circle definition of the sine and cosine

Q

y

functions to all six trigonometric functions. Let P be a point in QI

on the unit circle, so that the line segment OP in Figure 5.1.10

R

S

has length 1 and makes an acute angle θ with the positive x-axis.

P

Identify each of the six trigonometric functions of θ with exactly

one of the line segments in Figure 5.1.10, keeping in mind that the

radius of the circle is 1. To get you started, we have sin θ = MP

N

θ

x

(why?).

O

M

1

14. For Exercise 13, how would you draw the line segments in Fig-

ure 5.1.10 if θ was in QII? Recall that some of the trigonometric

functions are negative in QII, so you will have to come up with a

convention for how to treat some of the line segment lengths as

Figure 5.1.10

negative.

15. For any point ( x, y) on the unit circle and any angle α, show that the point Rα( x, y) defined by Rα( x, y) = ( x cos α − y sin α, x sin α + y cos α) is also on the unit circle. What is the geometric interpretation of Rα( x, y)? Also, show that R− α( Rα( x, y)) = ( x, y) and Rβ( Rα( x, y)) = Rα+ β( x, y).

Properties of Graphs of Trigonometric Functions • Section 5.2

109

5.2 Properties of Graphs of Trigonometric Functions

We saw in Section 5.1 how the graphs of the trigonometric functions repeat every 2 π radi-

ans. In this section we will discuss this and other properties of graphs, especially for the

sinusoidal functions (sine and cosine).

First, recall that the domain of a function f ( x) is the set of all numbers x for which the

function is defined. For example, the domain of f ( x) = sin x is the set of all real numbers,

whereas the domain of f ( x) = tan x is the set of all real numbers except x = ± π ,

,

,

2

± 3 π

2

± 5 π

2

.... The range of a function f ( x) is the set of all values that f ( x) can take over its domain.

For example, the range of f ( x) = sin x is the set of all real numbers between −1 and 1 (i.e.

the interval [−1,1]), whereas the range of f ( x) = tan x is the set of all real numbers, as we

can see from their graphs.

A function f ( x) is periodic if there exists a number p > 0 such that x + p is in the domain of f ( x) whenever x is, and if the following relation holds:

f ( x + p) = f ( x) for all x

(5.1)

There could be many numbers p that satisfy the above requirements. If there is a smallest

such number p, then we call that number the period of the function f ( x).

Example 5.3

The functions sin x, cos x, csc x, and sec x all have the same period: 2 π radians. We saw in Section 5.1 that the graphs of y = tan x and y = cot x repeat every 2 π radians but they also repeat every π

radians. Thus, the functions tan x and cot x have a period of π radians.

Example 5.4

What is the period of f ( x) = sin 2 x ?

Solution: The graph of y = sin 2 x is shown in Figure 5.2.1, along with the graph of y = sin x for comparison, over the interval [0, 2 π]. Note that sin 2 x “goes twice as fast” as sin x.

y

y = sin 2 x

1

y = sin x

0

x

π

π

3 π

π

5 π

3 π

7 π

2 π

4

2

4

4

2

4

−1

Figure 5.2.1

Graph of y = sin 2 x

For example, for x from 0 to π , sin x goes from 0 to 1, but sin 2 x is able to go from 0 to 1 quicker,

2

just over the interval [0, π ]. While sin x takes a full 2 π radians to go through an entire cycle (the

4

largest part of the graph that does not repeat), sin 2 x goes through an entire cycle in just π radians.

So the period of sin 2 x is π radians.

110

Chapter 5 • Graphing and Inverse Functions

§5.2

The above example made use of the graph of sin 2 x, but the period can be found analyt-

ically. Since sin x has period 2 π,1 we know that sin ( x + 2 π) = sin x for all x. Since 2 x is a number for all x, this means in particular that sin (2 x + 2 π) = sin 2 x for all x. Now define

f ( x) = sin 2 x. Then

f ( x + π) = sin 2( x + π)

= sin (2 x + 2 π)

= sin 2 x (as we showed above)

= f ( x)

for all x, so the period p of sin 2 x is at most π, by our definition of period. We have to show

that p > 0 can not be smaller than π. To do this, we will use a proof by contradiction. That

is, assume that 0 < p < π, then show that this leads to some contradiction, and hence can not

be true. So suppose 0 < p < π. Then 0 < 2 p < 2 π, and hence

sin 2 x = f ( x)

= f ( x + p) (since p is the period of f ( x))

= sin 2( x + p)

= sin (2 x + 2 p)

for all x. Since any number u can be written as 2 x for some x (i.e u = 2( u/2)), this means that sin u = sin ( u + 2 p) for all real numbers u, and hence the period of sin x is as most 2 p.

This is a contradiction. Why? Because the period of sin x is 2 π > 2 p. Hence, the period p of

sin 2 x can not be less than π, so the period must equal π.

The above may seem like a lot of work to prove something that was visually obvious from

the graph (and intuitively obvious by the “twice as fast” idea). Luckily, we do not need to

go through all that work for each function, since a similar argument works when sin 2 x is

replaced by sin ωx for any positive real number ω: instead of dividing 2 π by 2 to get the

period, divide by ω. And the argument works for the other trigonometric functions as well.

Thus, we get:

For any number ω > 0:

2 π

2 π

sin ωx has period

csc ωx has period

ω

ω

2 π

2 π

cos ωx has period

sec ωx has period

ω

ω

π

π

tan ωx has period

cot ωx has period

ω

ω

If ω < 0, then use sin (− A) = −sin A and cos (− A) = cos A (e.g. sin (−3 x) = −sin 3 x).

1We will usually leave out the “radians” part when discussing periods from now on.

Properties of Graphs of Trigonometric Functions • Section 5.2

111

Example 5.5

The period of y = cos 3 x is 2 π and the period of y

x is 4 π. The graphs of both functions are

3

= cos 12

shown in Figure 5.2.2:

y = cos 1 x

y

2

y = cos 3 x

1

0

x

π

π

π

2 π

5 π

π 7 π 4 π 3 π 5 π 11 π 2 π 13 π 7 π 5 π 8 π 17 π 3 π 19 π 10 π 7 π 11 π 23 π 4 π

6

3

2

3

6

6

3

2

3

6

6

3

2

3

6

6

3

2

3

6

−1

Figure 5.2.2

Graph of y = cos 3 x and y = cos 1 x

2

We know that −1 ≤ sin x ≤ 1 and −1 ≤ cos x ≤ 1 for all x. Thus, for a constant A = 0,

−| A| ≤ A sin x ≤ | A| and

− | A| ≤ A cos x ≤ | A|

for all x. In this case, we call | A| the amplitude of the functions y = A sin x and y = A cos x.

In general, the amplitude of a periodic curve f ( x) is half the difference of the largest and

smallest values that f ( x) can take:

(maximum of f ( x)) − (minimum of f ( x))

Amplitude of f ( x) =

2

In other words, the amplitude is the distance from either the top or bottom of the curve to

the horizontal line that divides the curve in half, as in Figure 5.2.3.

y

| A|

| A|

2 | A|

0

x

π

π

3 π

π

5 π

3 π

7 π

2 π

4

2

4

4

2

4

| A|

−| A|

Figure 5.2.3

Amplitude = max−min

2

= | A|−(−| A|)

2

= | A|

Not all periodic curves have an amplitude. For example, tan x has neither a maximum

nor a minimum, so its amplitude is undefined. Likewise, cot x, csc x, and sec x do not have

an amplitude. Since the amplitude involves vertical distances, it has no effect on the period

of a function, and vice versa.

112

Chapter 5 • Graphing and Inverse Functions

§5.2

Example 5.6

Find the amplitude and period of y = 3 cos 2 x.

Solution: The amplitude is |3| = 3 and the period is 2 π

2 = π. The graph is shown in Figure 5.2.4:

y

3

2

3

1

6

0

x

π

π

π

2 π

5 π

π

7 π

4 π

3 π

5 π

11 π

2 π

−1

6

3

2

3

6

6

3

2

3

6

3

−2

−3

Figure 5.2.4

y = 3 cos 2 x

Example 5.7

Find the amplitude and period of y = 2 − 3 sin 2 π x.

3

Solution: The amplitude of −3 sin 2 π x is

3

|−3| = 3. Adding 2 to that function to get the function

y = 2 − 3 sin 2 π x does not change the amplitude, even though it does change the maximum and

3

minimum. It just shifts the entire graph upward by 2. So in this case, we have

max − min

5 − (−1)

6

Amplitude =

=

=

= 3 .

2

2

2

2 π

The period is

= 3. The graph is shown in Figure 5.2.5:

2 π

3

y

5

4

3

3

6

2

1

3

0

x

3

3

9

3

4

2

4

−1

Figure 5.2.5

y = 2 − 3 sin 2 π x

3

Properties of Graphs of Trigonometric Functions • Section 5.2

113

Example 5.8

Find the amplitude and period of y = 2 sin ( x 2).

Solution: This is not a periodic function, since the angle that we are taking the sine of, x 2, is not a

linear function of x, i.e. is not of the form ax + b for some constants a and b. Recall how we argued that sin 2 x was “twice as fast” as sin x, so that its period was π instead of 2 π. Can we say that

sin ( x 2) is some constant times as fast as sin x ? No. In fact, we see that the “speed” of the curve keeps

increasing as x gets larger, since x 2 grows at a variable rate, not a constant rate. This can be seen in

the graph of y = 2 sin ( x 2), shown in Figure 5.2.6:2

2

1.5

1

0.5

y

0

-0.5

-1

-1.5

-2

0

π

π

3 π

2 π

2

2

x

Figure 5.2.6

y = 2 sin ( x 2)

Notice how the curve “speeds up” as x gets larger, making the “waves” narrower and narrower.

Thus, y = 2 sin ( x 2) has no period. Despite this, it appears that the function does have an amplitude,

namely 2. To see why, note that since |sin θ| ≤ 1 for all θ, we have

|2 sin ( x 2)| = |2| · |sin ( x 2)| ≤ 2 · 1 = 2 .

In the exercises you will be asked to find values of x such that 2 sin ( x 2) reaches the maximum value

2 and the minimum value −2. Thus, the amplitude is indeed 2.

Note: This curve is still sinusoidal despite not being periodic, since the general shape is still that of

a “sine wave”, albeit one with variable cycles.

So far in our examples we have been able to determine the amplitudes of sinusoidal curves

fairly easily. This will not always be the case.

2This graph was created using Gnuplot, an open-source graphing program which is freely available at http://

gnuplot.info. See Appendix B for a brief tutorial on how to use Gnuplot.

114

Chapter 5 • Graphing and Inverse Functions

§5.2

Example 5.9

Find the amplitude and period of y = 3 sin x + 4 cos x.

Solution: This is sometimes called a combination sinusoidal curve, since it is the sum of two such

curves. The period is still simple to determine: since sin x and cos x each repeat every 2 π radians,

then so does the combination 3 sin x + 4 cos x. Thus, y = 3 sin x + 4 cos x has period 2 π. We can see this in the graph, shown in Figure 5.2.7:

5

4

3

2

1

y

0

-1

-2

-3

-4

-5

0

π

π

3 π

2 π

5 π

3 π

7 π

4 π

2

2

2

2

x

Figure 5.2.7

y = 3 sin x + 4 cos x

The graph suggests that the amplitude is 5, which may not be immediately obvious just by looking

at how the function is defined. In fact, the definition y = 3 sin x +4 cos x may tempt you to think that

the amplitude is 7, since the largest that 3 sin x could be is 3 and the largest that 4 cos x could be is

4, so that the largest their sum could be is 3 + 4 = 7. However, 3 sin x can never equal 3 for the same

x that makes 4 cos x equal to 4 (why?).

There is a useful technique (which we will discuss further in Chapter 6) for showing

that the amplitude of y = 3 sin x + 4 cos x is 5. Let θ be the angle shown in the right

5

triangle in Figure 5.2.8. Then cos θ = 3 and sin θ = 4 . We can use this as follows:

4

5

5

y = 3 sin x + 4 cos x

θ

3

= 5 3 sin x

cos x

5

+ 45

Figure 5.2.8

= 5(cos θ sin x + sin θ cos x)

= 5 sin ( x + θ) (by the sine addition formula)

Thus, | y| = |5 sin ( x + θ)| = |5| · |sin ( x + θ)| ≤ (5)(1) = 5, so the amplitude of y = 3 sin x + 4 cos x is 5.

Properties of Graphs of Trigonometric Functions • Section 5.2

115

In general, a combination of sines and cosines will have a period equal to the lowest com-

mon multiple of the periods of the sines and cosines being added. In Example 5.9, sin x

and cos x each have period 2 π, so the lowest common multiple (which is always an integer

multiple) is 1 · 2 π = 2 π.

Example 5.10

Find the period of y = cos 6 x + sin 4 x.

Solution: The period of cos 6 x is 2 π

, and the period of sin 4 x is 2 π

. The lowest common

6 = π

3

4 = π

2

multiple of π and π is π:

3

2

1 · π

1

3 = π

3

· π 2 = π 2

2 · π

2

3 = 2 π

3

· π 2 = π

3 · π 3 = π

Thus, the period of y = cos 6 x + sin 4 x is π. We can see this from its graph in Figure 5.2.9:

2

1.5

1

0.5

y

0

-0.5

-1

-1.5

-2

0

π

π

3 π

2 π

2

2

x

Figure 5.2.9

y = cos 6 x + sin 4 x

What about the amplitude? Unfortunately we can not use the technique from Example 5.9, since

we are not taking the cosine and sine of the same angle; we are taking the cosine of 6 x but the sine of

4 x. In this case, it appears from the graph that the maximum is close to 2 and the minimum is close

to −2. In Chapter 6, we will describe how to use a numerical computation program to show that the

maximum and minimum are ±1.90596111871578, respectively (accurate to within ≈ 2.2204×10−16).

Hence, the amplitude is 1.90596111871578.

116

Chapter 5 • Graphing and Inverse Functions

§5.2

Generalizing Example 5.9, an expression of the form a sin ωx + b cos ωx is equivalent to

a 2 + b 2 sin ( x + θ), where θ is an angle such that cos θ =

a

and sin θ =

b

. So

a 2+ b 2

a 2+ b 2

y = a sin ωx + b cos ωx will have amplitude

a 2 + b 2. Note that this method only works

when the angle ωx is the same in both the sine and cosine terms.

We have seen how adding a constant to a function shifts the entire graph vertically. We

will now see how to shift the entire graph of a periodic curve horizontally.

Consider a function of the form y

y

= A sin ωx, where A

and ω are nonzero constants. For simplicity we will assume

period = 2 π

ω

that A > 0 and ω > 0 (in general either one could be nega-

A

tive). Then the amplitude is A and the period is 2 π . The

ω

graph is shown in Figure 5.2.10.

0

x

Now consider the function y = A sin ( ωx − φ), where φ is

π

2 π

ω

ω

some constant. The amplitude is still A, and the period is

still 2 π , since ωx

ω

− φ is a linear function of x. Also, we know

− A

that the sine function goes through an entire cycle when its

Figure 5.2.10 y = A sin ωx

angle goes from 0 to 2 π. Here, we are taking the sine of the

angle ωx − φ. So as ωx − φ goes from 0 to 2 π, an entire cycle of the function y = A sin ( ωx − φ) will be traced out. That cycle starts when

φ

ωx − φ = 0

⇒

x = ω

and ends when

2 π

φ

ωx − φ = 2 π

⇒

x =

+

.

ω

ω

Thus, the graph of y = A sin ( ωx − φ) is just the graph of y = A sin ωx shifted horizontally by φ , as in Figure 5.2.11. The graph is shifted to the right when φ

ω

> 0, and to the left when

φ < 0. The amount φ of the shift is called the phase shift of the graph.

ω

y

y

period = 2 π

period

ω

= 2 π

ω

A

A

0

x

x

φ

0

2 π

φ

φ

2 π

φ

ω

ω + ω

ω

ω + ω

− A phase shift

− A

phase shift

(a) φ > 0: right shift

(b) φ < 0: left shift

Figure 5.2.11

Phase shift for y = A sin ( ωx − φ)

Properties of Graphs of Trigonometric Functions • Section 5.2

117

The phase shift is defined similarly for the other trigonometric functions.

Example 5.11

Find the amplitude, period, and phase shift of y = 3 cos (2 x − π).

Solution: The amplitude is 3, the period is 2 π

. The graph is shown in

2 = π, and the phase shift is π

2

Figure 5.2.12:

y

period = π

3

2

amplitude = 3

1

0

x

π

π

3 π

2 π

−1

2

2

−2

−3

phase shift = π 2

Figure 5.2.12

y = 3 cos (2 x − π)

Notice that the graph is the same as the graph of y = 3 cos 2 x shifted to the right by π , the amount

2

of the phase shift.

Example 5.12

Find the amplitude, period, and phase shift of y = −2 sin 3 x + π .

2

− π

Solution: The amplitude is 2, the period is 2 π , and the phase shift is

2

. Notice the negative

3

3 = − π

6

sign in the phase shift, since 3 x + π = 3 x − (− π) is in the form ωx − φ. The graph is shown in Figure

5.2.13:

y

period = 2 π

3

2

amplitude = 2

1

x

0

− π

π

π

π

2 π

5 π

π

7 π

4 π

6

6

3

2

3

6

6

3

−1

−2

phase shift = − π 6

Figure 5.2.13

y = −2 sin 3 x + π 2

118

Chapter 5 • Graphing and Inverse Functions

§5.2

In engineering two periodic functions with the same period are said to be out of phase if

their phase shifts differ. For example, sin x − π and sin x would be π radians (or 30◦) out of

6

6

phase, and sin x would be said to lag sin x − π by π radians, while sin x

leads sin x by

6

6

− π 6

π radians. Periodic functions with the same period and the same phase shift are in phase.

6

The following is a summary of the properties of trigonometric graphs:

For any constants A = 0, ω = 0, and φ:

y = A sin ( ωx − φ) has amplitude | A|, period 2 π , and phase shift φ

ω

ω

y = A cos ( ωx − φ) has amplitude | A|, period 2 π , and phase shift φ

ω

ω

y = A tan ( ωx − φ) has undefined amplitude, period π , and phase shift φ

ω

ω

y = A csc ( ωx − φ) has undefined amplitude, period 2 π , and phase shift φ

ω

ω

y = A sec ( ωx − φ) has undefined amplitude, period 2 π , and phase shift φ

ω

ω

y = A cot ( ωx − φ) has undefined amplitude, period π , and phase shift φ

ω

ω

Exercises

For Exercises 1-12, find the amplitude, period, and phase shift of the given function. Then graph one

cycle of the function, either by hand or by using Gnuplot (see Appendix B).

1. y = 3 cos πx

2. y = sin (2 πx − π)

3. y = −sin (5 x + 3)

4. y = 1 + 8 cos (6 x − π)

5. y = 2 + cos (5 x + π)

6. y = 1 − sin (3 π − 2 x)

7. y = 1 − cos (3 π − 2 x)

8. y = 2 tan ( x − 1)

9. y = 1 − tan (3 π − 2 x)

10. y = sec (2 x + 1)

11. y = 2csc (2 x − 1)

12. y = 2 + 4 cot (1 − x)

13. For the function y = 2 sin ( x 2) in Example 5.8, for which values of x does the function reach its maximum value 2, and for which values of x does it reach its minimum value −2?

14. For the function y = 3 sin x+4 cos x in Example 5.9, for which values of x does the function reach its maximum value 5, and for which values of x does it reach its minimum value −5? You can

restrict your answers to be between 0 and 2 π.

15. Graph the function y = sin2 x from x = 0 to x = 2 π, either by hand or by using Gnuplot. What are the amplitude and period of this function?

16. The current i( t) in an AC electrical circuit at time t ≥ 0 is given by i( t) = Im sin ωt, and the voltage v( t) is given by v( t) = Vm sin ωt, where Vm > Im > 0 and ω > 0 are constants. Sketch one cycle of both i( t) and v( t) together on the same graph (i.e. on the same set of axes). Are the current and voltage in phase or out of phase?

17. Repeat Exercise 16 with i( t) the same as before but with v( t) = Vm sin ωt + π .

4

18. Repeat Exercise 16 with i( t) = − Im cos ωt − π and v( t)

.

3

= Vm sin ωt − 5 π

6

For Exercises 19-21, find the amplitude and period of the given function. Then graph one cycle of the

function, either by hand or by using Gnuplot.

Properties of Graphs of Trigonometric Functions • Section 5.2

119

19. y = 3 sin πx − 5 cos πx

20. y = −5 sin 3 x + 12 cos 3 x

21. y = 2 cos x + 2 sin x

22. Find the amplitude of the function y = 2 sin ( x 2) + cos ( x 2).

For Exercises 23-25, find the period of the given function. Graph one cycle using Gnuplot.

23. y = sin 3 x − cos 5 x

24. y = sin x

25. y

x

3 + 2 cos 3 x

4

= 2 sin πx + 3 cos π 3

26. Let y = 0.5 sin x sin 12 x . Its graph for x from 0 to 4 π is shown in Figure 5.2.14:

1

0.5 ∗ sin( x) ∗ sin(12 ∗ x)

0.5 ∗ sin( x)

−0.5 ∗ sin( x)

0.5

y

0

-0.5

-1

0

π

2 π

3 π

4 π

x

Figure 5.2.14

Modulated wave y = 0.5 sin x sin 12 x

You can think of this function as sin 12 x with a sinusoidally varying “amplitude”of 0.5 sin x. What

is the period of this function? From the graph it looks like the amplitude may be 0.5. Without

finding the exact amplitude, explain why the amplitude is in fact less than 0.5. The function above

is known as a modulated wave, and the functions ±0.5 sin x form an amplitude envelope for the

wave (i.e. they enclose the wave). Use an identity from Section 3.4 to write this function as a sum

of sinusoidal curves.

27. Use Gnuplot to graph the function y = x 2 sin 10 x from x = −2 π to x = 2 π. What functions form its amplitude envelope? (Note: Use set samples 500 in Gnuplot.)

28. Use Gnuplot to graph the function y = 1 sin 80 x from x = 0.2 to x = π. What functions form its x 2

amplitude envelope? (Note: Use set samples 500 in Gnuplot.)

29. Does the function y = sin πx + cos x have a period? Explain your answer.

30. Use Gnuplot to graph the function y = sin x from x

x

= −4 π to x = 4 π. What happens at x = 0?

120

Chapter 5 • Graphing and Inverse Functions

§5.3

5.3 Inverse Trigonometric Functions

We have briefly mentioned the inverse trigonometric functions before, for example in Section

✄

✄

✄

1.3 when we discussed how to use the sin−1 , cos−1 , and tan−1 buttons on a calculator to find

✂

✁ ✂

✁

✂

✁

an angle that has a certain trigonometric function value. We will now define those inverse

functions and determine their graphs.

Domain

Range

Recall that a function is a rule that assigns a single object

y from one set (the range) to each object x from another set

f

(the domain). We can write that rule as y

x

y

= f ( x), where f is

y = f ( x)

the function (see Figure 5.3.1). There is a simple vertical rule

Figure 5.3.1

for determining whether a rule y = f ( x) is a function: f is a

function if and only if every vertical line intersects the graph

of y = f ( x) in the xy-coordinate plane at most once (see Figure 5.3.2).

y

y

y = f ( x)

y = f ( x)

x

x

(a) f is a function

(b) f is not a function

Figure 5.3.2

Vertical rule for functions

Recall that a function f is one-to-one (often written as 1 − 1) if it assigns distinct values

of y to distinct values of x. In other words, if x 1 = x 2 then f ( x 1) = f ( x 2). Equivalently, f is one-to-one if f ( x 1) = f ( x 2) implies x 1 = x 2. There is a simple horizontal rule for determining whether a function y = f ( x) is one-to-one: f is one-to-one if and only if every horizontal line

intersects the graph of y = f ( x) in the xy-coordinate plane at most once (see Figure 5.3.3).

y

y

y = f ( x)

y = f ( x)

x

x

(a) f is one-to-one

(b) f is not one-to-one

Figure 5.3.3

Horizontal rule for one-to-one functions

If a function f is one-to-one on its domain, then f has an inverse function, denoted by

f −1, such that y = f ( x) if and only if f −1( y) = x. The domain of f −1 is the range of f .

Inverse Trigonometric Functions • Section 5.3

121

The basic idea is that f −1 “undoes” what f does, and vice versa. In other words,

f −1( f ( x)) = x for all x in the domain of f , and

f ( f −1( y)) = y for all y in the range of f .

We know from their graphs that none of the trigonometric functions are one-to-one over

their entire domains. However, we can restrict those functions to subsets of their domains

where they are one-to-one. For example, y = sin x is one-to-one over the interval − π , π , as

2 2

we see in the graph below:

y

y = sin x

1

x

0

π

− π

π

− π 2

2

−1

Figure 5.3.4

y = sin x with x restricted to − π , π

2 2

For − π

we have

2 ≤ x ≤ π

2

−1 ≤ sin x ≤ 1, so we can define the inverse sine function y =

sin−1 x (sometimes called the arc sine and denoted by y = arcsin x) whose domain is the

interval [−1,1] and whose range is the interval − π , π . In other words:

2 2

sin−1(sin y) = y for − π

(5.2)

2 ≤ y ≤ π

2

sin (sin−1 x) = x for −1 ≤ x ≤ 1

(5.3)

Example 5.13

Find sin−1 sin π .

4

π

Solution: Since − π

, we know that sin−1 sin π

, by formula (5.2).

2 ≤ π

4 ≤ π

2

4

= 4

Example 5.14

Find sin−1 sin 5 π .

4

Solution: Since 5 π

, we can not use formula (5.2). But we know that sin 5 π

. Thus,

4 > π

2

4 = − 12

sin−1 sin 5 π

is, by definition, the angle y such that

and sin y

. That

4

= sin−1 − 1

− π

= − 1

2

2 ≤ y ≤ π

2

2

angle is y = − π , since

4

sin − π

.

4

= −sin π 4 = − 12

Thus, sin−1 sin 5 π

.

4

= − π 4

122

Chapter 5 • Graphing and Inverse Functions

§5.3

Example 5.14 illustrates an important point: sin−1 x should always be a number between

− π and π . If you get a number outside that range, then you made a mistake somewhere.

2

2

✄

This why in Example 1.27 in Section 1.5 we got sin−1(−0.682) = −43◦ when using the sin−1

✂

✁

button on a calculator. Instead of an angle between 0◦ and 360◦ (i.e. 0 to 2 π radians) we got

an angle between −90◦ and 90◦ (i.e. − π to π radians).

2

2

In general, the graph of an inverse function f −1 is the reflection of the graph of f around

the line y = x. The graph of y = sin−1 x is shown in Figure 5.3.5. Notice the symmetry about the line y = x with the graph of y = sin x.

y

y = sin−1 x

π

2

1

y = sin x

x

0

π

−1

1

− π 2

2

−1

y = x

− π 2

Figure 5.3.5

Graph of y = sin−1 x

The inverse cosine function y = cos−1 x (sometimes called the arc cosine and denoted

by y = arccos x) can be determined in a similar fashion. The function y = cos x is one-to-one

over the interval [0, π], as we see in the graph below:

y

y = cos x

1

x

0

π

π

− π

3 π

2

2

2

−1

Figure 5.3.6

y = cos x with x restricted to [0, π]

Thus, y = cos−1 x is a function whose domain is the interval [−1,1] and whose range is the

interval [0, π]. In other words:

Inverse Trigonometric Functions • Section 5.3

123

cos−1(cos y) = y for 0 ≤ y ≤ π

(5.4)

cos (cos−1 x) = x for −1 ≤ x ≤ 1

(5.5)

The graph of y = cos−1 x is shown below in Figure 5.3.7. Notice the symmetry about the

line y = x with the graph of y = cos x.

y

y = cos−1 x

π

1

y = cos x

x

0

π

−1

1

π

− π 2

2

y = x

−1

Figure 5.3.7

Graph of y = cos−1 x

Example 5.15

Find cos−1 cos π .

3

π

Solution: Since 0 ≤ π

, by formula (5.4).

3 ≤ π, we know that cos−1 cos π

3

= 3

Example 5.16

Find cos−1 cos 4 π .

3

Solution: Since 4 π

. Thus,

3 > π, we can not use formula (5.4).

But we know that cos 4 π

3 = − 1

2

cos−1 cos 4 π

is, by definition, the angle y such that 0

. That

3

= cos−1 − 12

≤ y ≤ π and cos y = − 12

angle is y = 2 π (i.e. 120◦). Thus, cos−1 cos 4 π

.

3

3

= 2 π

3

Examples 5.14 and 5.16 may be confusing, since they seem to violate the general rule for inverse functions that f −1( f ( x)) = x for all x in the domain of f . But that rule only applies when the function f is one-to-one over its entire domain. We had to restrict the sine and

cosine functions to very small subsets of their entire domains in order for those functions

to be one-to-one. That general rule, therefore, only holds for x in those small subsets in the

case of the inverse sine and inverse cosine.

124

Chapter 5 • Graphing and Inverse Functions

§5.3

The inverse tangent function y = tan−1 x (sometimes called the arc tangent and de-

noted by y = arctan x) can be determined similarly. The function y = tan x is one-to-one over

the interval − π , π , as we see in Figure 5.3.8:

2 2

y

3

2

1

x

0

π

π

− π 2

− π 4

4

2

y = tan x

−1

−2

−3

Figure 5.3.8

y = tan x with x restricted to − π , π

2 2

The graph of y = tan−1 x is shown below in Figure 5.3.9. Notice that the vertical asymp-

totes for y = tan x become horizontal asymptotes for y = tan−1 x. Note also the symmetry

about the line y = x with the graph of y = tan x.

y

3

y = tan x

2

π

2

1

y = tan−1 x

x

0

π

π

− π 2

− π 4

4

2

−1

− π 2

y = x

−2

−3

Figure 5.3.9

Graph of y = tan−1 x

Inverse Trigonometric Functions • Section 5.3

125

Thus, y = tan−1 x is a function whose domain is the set of all real numbers and whose

range is the interval − π , π . In other words:

2 2

tan−1(tan y) = y for − π

(5.6)

2 < y < π

2

tan (tan−1 x) = x for all real x

(5.7)

Example 5.17

Find tan−1 tan π .

4

π

Solution: Since − π

, we know that tan−1 tan π

, by formula (5.6).

2 ≤ π

4 ≤ π

2

4

= 4

Example 5.18

Find tan−1 (tan π).

Solution: Since π > π , we can not use formula (5.6). But we know that tan π

2

= 0. Thus, tan−1 (tan π) =

tan−1 0 is, by definition, the angle y such that − π

and tan y

2 ≤ y ≤ π

2

= 0. That angle is y = 0. Thus,

tan−1 (tan π) = 0 .

Example 5.19

Find the exact value of cos sin−1 − 1 .

4

Solution: Let θ = sin−1 − 1 . We know that

, so since sin θ

4

− π 2 ≤ θ ≤ π 2

= − 14 < 0, θ must be in QIV.

Hence cos θ > 0. Thus,

1 2

15

15

cos2 θ = 1 − sin2 θ = 1 − −

=

⇒

cos θ =

.

4

16

4

15

Note that we took the positive square root above since cos θ > 0. Thus, cos sin−1 − 1

.

4

=

4

Example 5.20

x

Show that tan (sin−1 x) =

for −1 < x < 1.

1 − x 2

Solution: When x = 0, the formula holds trivially, since

0

1

tan (sin−1 0) = tan 0 = 0 =

.

x

1 − 02

Now suppose that 0 < x < 1. Let θ = sin−1 x. Then θ is in QI and sin θ = x.

θ

Draw a right triangle with an angle θ such that the opposite leg has length x and

1 − x 2

the hypotenuse has length 1, as in Figure 5.3.10 (note that this is possible since

Figure 5.3.10

0 < x < 1). Then sin θ = x 1 = x. By the Pythagorean Theorem, the adjacent leg has

length

1 − x 2. Thus, tan θ =

x

.

1− x 2

If −1 < x < 0 then θ = sin−1 x is in QIV. So we can draw the same triangle except that it would be

“upside down” and we would again have tan θ =

x

, since the tangent and sine have the same sign

1− x 2

x

(negative) in QIV. Thus, tan (sin−1 x) =

for −1 < x < 1.

1 − x 2

126

Chapter 5 • Graphing and Inverse Functions

§5.3

The inverse functions for cotangent, cosecant, and secant can be determined by looking at

their graphs. For example, the function y = cot x is one-to-one in the interval (0, π), where it

has a range equal to the set of all real numbers. Thus, the inverse cotangent y = cot−1 x is

a function whose domain is the set of all real numbers and whose range is the interval (0, π).

In other words:

cot−1(cot y) = y for 0 < y < π

(5.8)

cot (cot−1 x) = x for all real x

(5.9)

The graph of y = cot−1 x is shown below in Figure 5.3.11.

y

π

π

2

y = cot−1 x

x

0

π

π

3 π

− 3 π

− π 2

− π 4

4

2

4

4

Figure 5.3.11

Graph of y = cot−1 x

Similarly, it can be shown that the inverse cosecant y = csc−1 x is a function whose

domain is | x| ≥ 1 and whose range is − π

, y

2 ≤ y ≤ π

2

= 0. Likewise, the inverse secant

y = sec−1 x is a function whose domain is | x| ≥ 1 and whose range is 0 ≤ y ≤ π, y = π .

2

π

π

csc−1(csc y) = y for − ≤ y ≤ , y = 0

(5.10)

2

2

csc (csc−1 x) = x for | x| ≥ 1

(5.11)

π

sec−1(sec y) = y for 0 ≤ y ≤ π, y =

(5.12)

2

sec (sec−1 x) = x for | x| ≥ 1

(5.13)

It is also common to call cot−1 x, csc−1 x, and sec−1 x the arc cotangent, arc cosecant,

and arc secant, respectively, of x. The graphs of y = csc−1 x and y = sec−1 x are shown in

Figure 5.3.12:

Inverse Trigonometric Functions • Section 5.3

127

y

y

π

π

2

y = csc−1 x

y = sec−1 x

π

x

2

0

−1

1

x

− π 2

0

−1

1

(a) Graph of y = csc−1 x

(b) Graph of y = sec−1 x

Figure 5.3.12

Example 5.21

Prove the identity tan−1 x + cot−1 x = π .

2

Solution: Let θ = cot−1 x. Using relations from Section 1.5, we have

tan π 2 − θ = −tan θ − π 2 = cot θ = cot (cot−1 x) = x ,

by formula (5.9). So since tan (tan−1 x) = x for all x, this means that tan (tan−1 x) = tan π 2 − θ .

Thus, tan (tan−1 x) = tan π 2 − cot−1 x . Now, we know that 0 < cot−1 x < π, so − π 2 < π 2 − cot−1 x < π , i.e. π

2

2 − cot−1 x is in the restricted subset on which the tangent function is one-to-one. Hence,

tan (tan−1 x) = tan π 2 − cot−1 x implies that tan−1 x = π 2 − cot−1 x, which proves the identity.

Example 5.22

a + b

Is tan−1 a + tan−1 b = tan−1

an identity?

1 − ab

tan A + tan B

Solution: In the tangent addition formula tan ( A + B) =

, let A = tan−1 a and B =

1 − tan A tan B

tan−1 b. Then

tan (tan−1 a) + tan (tan−1 b)

tan (tan−1 a + tan−1 b) = 1 − tan(tan−1 a) tan(tan−1 b)

a + b

=

by formula (5.7), so it seems that we have

1 − ab

tan−1 a + tan−1 b = tan−1 a + b

1 − ab

by definition of the inverse tangent. However, recall that − π

for all real numbers x. So

2 < tan−1 x < π

2

in particular, we must have − π

. But it is possible that tan−1 a

2 < tan−1

a+ b

1− ab < π

2

+ tan−1 b is not in

the interval − π , π . For example,

2 2

tan−1 1 + tan−1 2 = 1.892547 > π 2 ≈ 1.570796 .

And we see that tan−1

1+2

1−(1)(2) = tan−1(−3) = −1.249045 = tan−1 1 + tan−1 2. So the formula is only

true when − π

.

2 < tan−1 a + tan−1 b < π

2

128

Chapter 5 • Graphing and Inverse Functions

§5.3

Exercises

For Exercises 1-25, find the exact value of the given expression in radians.

1. tan−1 1

2. tan−1 (−1)

3. tan−1 0

4. cos−1 1

5. cos−1 (−1)

6. cos−1 0

7. sin−1 1

8. sin−1 (−1)

9. sin−1 0

10. sin−1 sin π 3

11. sin−1 sin 4 π

12. sin−1 sin

13. cos−1 cos π

14. cos−1 cos

3

− 5 π

6

7

− π

10

15. cos−1 cos 6 π

16. tan−1 tan 4 π

17. tan−1 tan

18. cot−1 cot 4 π

5

3

− 5 π

6

3

19. csc−1 csc − π

20. sec−1 sec 6 π

21. cos sin−1

5

22. cos sin−1

9

5

13

− 45

23. sin−1 3

24. sin−1 5

25. tan−1 3

5 + sin−1 4

5

13 + cos−1 5

13

5 + cot−1 3

5

For Exercises 26-33, prove the given identity.

26. cos (sin−1 x) =

1 − x 2

27. sin (cos−1 x) =

1 − x 2

28. sin−1 x + cos−1 x = π

29. sec−1 x

2

+ csc−1 x = π 2

30. sin−1(− x) = −sin−1 x

31. cos−1(− x) + cos−1 x = π

32. cot−1 x = tan−1 1 for x

for x

x

> 0

33. tan−1 x + tan−1 1 x = π 2

> 0

a + b

34. In Example 5.22 we showed that the formula tan−1 a + tan−1 b = tan−1

does not always

1 − ab

a + b

hold. Does the formula tan (tan−1 a + tan−1 b) =

, which was part of that example, always

1 − ab

hold? Explain your answer.

35. Show that tan−1 1

.

3 + tan−1 1

5 = tan−1 4

7

36. Show that tan−1 1

.

4 + tan−1 2

9 = tan−1 1

2

37. Figure 5.3.13 shows three equal squares lined up against each other. For the angles α, β, and γ

in the picture, show that α = β + γ. ( Hint: Consider the tangents of the angles. )

γ

β

α

Figure 5.3.13

Exercise 37

38. Sketch the graph of y = sin−1 2 x.

39. Write a computer program to solve a triangle in the case where you are given three sides. Your

program should read in the three sides as input parameters and print the three angles in degrees

as output if a solution exists. Note that since most computer languages use radians for their

inverse trigonometric functions, you will likely have to do the conversion from radians to degrees

yourself in the program.

6 Additional Topics

6.1 Solving Trigonometric Equations

An equation involving trigonometric functions is called a trigonometric equation. For exam-

ple, an equation like

tan A = 0.75 ,

which we encountered in Chapter 1, is a trigonometric equation. In Chapter 1 we were

concerned only with finding a single solution (say, between 0◦ and 90◦). In this section we

will be concerned with finding the most general solution to such equations. ✄

To see what that means, take the above equation tan A = 0.75. Using the tan−1 calculator

✂

✁

button in degree mode, we get A = 36.87◦. However, we know that the tangent function

has period π rad = 180◦, i.e. it repeats every 180◦. Thus, there are many other possible

answers for the value of A, namely 36.87◦ + 180◦, 36.87◦ − 180◦, 36.87◦ + 360◦, 36.87◦ − 360◦,

36.87◦ + 540◦, etc. We can write this in a more compact form:

A = 36.87◦ + 180◦ k for k = 0, ±1, ±2, ...

This is the most general solution to the equation. Often the part that says “for k = 0, ±1,

±2, ...” is omitted since it is usually understood that k varies over all integers. The general

solution in radians would be:

A = 0.6435 + πk for k = 0, ±1, ±2, ...

Example 6.1

Solve the equation 2 sin θ + 1 = 0.

✄

Solution: Isolating sin θ gives sin θ = − 1 . Using the sin−1 calculator button in degree mode gives

2

✂

✁

us θ = −30◦, which is in QIV. Recall that the reflection of this angle around the y-axis into QIII also

has the same sine. That is, sin 210◦ = − 1 . Thus, since the sine function has period 2 π rad

2

= 360◦,

and since −30◦ does not differ from 210◦ by an integer multiple of 360◦, the general solution is:

θ = −30◦ + 360◦ k and 210◦ + 360◦ k

for k = 0, ±1, ±2, ...

In radians, the solution is:

π

7 π

θ = −

+ 2 πk and

+ 2 πk

for k = 0, ±1, ±2, ...

6

6

For the rest of this section we will write our solutions in radians.

129

130

Chapter 6 • Additional Topics

§6.1

Example 6.2

Solve the equation 2 cos2 θ − 1 = 0.

Solution: Isolating cos2 θ gives us

1

1

π

3 π

5 π

7 π

cos2 θ =

⇒

cos θ = ±

⇒

θ =

,

,

,

,

2

2

4

4

4

4

and since the period of cosine is 2 π, we would add 2 πk to each of those angles to get the general

solution. But notice that the above angles differ by multiples of π . So since every multiple of 2 π is

2

also a multiple of π , we can combine those four separate answers into one:

2

π

π

θ =

+

k

for k = 0, ±1, ±2, ...

4

2

Example 6.3

Solve the equation 2 sec θ = 1.

Solution: Isolating sec θ gives us

1

1

sec θ =

⇒

cos θ =

= 2 ,

2

sec θ

which is impossible. Thus, there is no solution .

Example 6.4

Solve the equation cos θ = tan θ.

Solution: The idea here is to use identities to put everything in terms of a single trigonometric

function:

cos θ = tan θ

sin θ

cos θ = cos θ

cos2 θ = sin θ

1 − sin2 θ = sin θ

0 = sin2 θ + sin θ − 1

The last equation looks more complicated than the original equation, but notice that it is actually a

quadratic equation: making the substitution x = sin θ, we have

−1 ±

1 − (4)(−1)

−1 ±

5

x 2 + x − 1 = 0

⇒

x =

=

= −1.618 , 0.618

2 (1)

2

by the quadratic formula from elementary algebra. But −1.618 < −1, so it is impossible that sin θ =

x = −1.618. Thus, we must have sin θ = x = 0.618. Hence, there are two possible solutions: θ = 0.666

rad in QI and its reflection π− θ = 2.475 rad around the y-axis in QII. Adding multiples of 2 π to these

gives us the general solution:

θ = 0.666 + 2 πk and 2.475 + 2 πk

for k = 0, ±1, ±2, ...

Solving Trigonometric Equations • Section 6.1

131

Example 6.5

Solve the equation sin θ = tan θ.

Solution: Trying the same method as in the previous example, we get

sin θ = tan θ

sin θ

sin θ = cos θ

sin θ cos θ = sin θ

sin θ cos θ − sin θ = 0

sin θ (cos θ − 1) = 0

⇒

sin θ = 0 or cos θ = 1

⇒

θ = 0 , π or θ = 0

⇒

θ = 0 , π ,

plus multiples of 2 π. So since the above angles are multiples of π, and every multiple of 2 π is a

multiple of π, we can combine the two answers into one for the general solution:

θ = πk

for k = 0, ±1, ±2, ...

Example 6.6

Solve the equation cos 3 θ = 1 .

2

Solution: The idea here is to solve for 3 θ first, using the most general solution, and then divide

that solution by 3. So since cos−1 1

, there are two possible solutions for 3 θ: 3 θ

in QI and its

2 = π

3

= π 3

reflection −3 θ = − π around the x-axis in QIV. Adding multiples of 2 π to these gives us:

3

π

3 θ = ±

+ 2 πk

for k = 0, ±1, ±2, ...

3

So dividing everything by 3 we get the general solution for θ:

π

2 π

θ = ±

+

k

for k = 0, ±1, ±2, ...

9

3

Example 6.7

Solve the equation sin 2 θ = sin θ.

Solution: Here we use the double-angle formula for sine:

sin 2 θ = sin θ

2 sin θ cos θ = sin θ

sin θ (2 cos θ − 1) = 0

1

⇒

sin θ = 0 or cos θ = 2

π

⇒

θ = 0 , π or θ = ± 3

π

⇒

θ = πk and

±

+ 2 πk

for k = 0, ±1, ±2, ...

3

132

Chapter 6 • Additional Topics

§6.1

Example 6.8

Solve the equation 2 sin θ − 3 cos θ = 1.

Solution: We will use the technique which we discussed in Chapter 5 for finding the

13

3

amplitude of a combination of sine and cosine functions. Take the coefficients 2 and

3 of sin θ and −cos θ, respectively, in the above equation and make them the legs

φ

of a right triangle, as in Figure 6.1.1. Let φ be the angle shown in the right triangle.

2

The leg with length 3 > 0 means that the angle φ is above the x-axis, and the leg with

length 2 > 0 means that φ is to the right of the y-axis. Hence, φ must be in QI. The

Figure 6.1.1

hypotenuse has length

13 by the Pythagorean Theorem, and hence cos φ = 2 and

13

sin θ = 3 . We can use this to transform the equation to solve as follows:

13

2 sin θ − 3 cos θ = 1

13

2

sin θ − 3 cos θ = 1

13

13

13 (cos φ sin θ − sin φ cos θ) = 1

13 sin ( θ − φ) = 1 (by the sine subtraction formula)

sin ( θ − φ) = 113

⇒

θ − φ = 0.281 or θ − φ = π − 0.281 = 2.861

⇒

θ = φ + 0.281 or θ = φ + 2.861

Now, since cos φ = 2 and φ is in QI, the most general solution for φ is φ = 0.983+2 πk for k = 0, ±1, 13

±2, ... . So since we needed to add multiples of 2 π to the solutions 0.281 and 2.861 anyway, the most

general solution for θ is:

θ = 0.983 + 0.281 + 2 πk and 0.983 + 2.861 + 2 πk

⇒

θ = 1.264 + 2 πk and 3.844 + 2 πk

for k = 0, ±1, ±2, ...

Note: In Example 6.8 if the equation had been 2 sin θ + 3 cos θ = 1 then we still would

have used a right triangle with legs of lengths 2 and 3, but we would have used the sine

addition formula instead of the subtraction formula.

Exercises

For Exercises 1-12, solve the given equation (in radians).

1. tan θ + 1 = 0

2. 2 cos θ + 1 = 0

3. sin 5 θ + 1 = 0

4. 2 cos2 θ − sin2 θ = 1

5. 2 sin2 θ − cos 2 θ = 0

6. 2 cos2 θ + 3 sin θ = 0

7. cos2 θ + 2 sin θ = −1

8. tan θ + cot θ = 2

9. sin θ = cos θ

10. 2 sin θ − 3 cos θ = 0

11. cos2 3 θ − 5 cos 3 θ + 4 = 0 12. 3 sin θ − 4 cos θ = 1

Numerical Methods in Trigonometry • Section 6.2

133

6.2 Numerical Methods in Trigonometry

We were able to solve the trigonometric equations in the previous section fairly easily, which

in general is not the case. For example, consider the equation

cos x = x .

(6.1)

There is a solution, as shown in Figure 6.2.1 below. The graphs of y = cos x and y = x intersect somewhere between x = 0 and x = 1, which means that there is an x in the interval

[0, 1] such that cos x = x.

4

cos( x)

x

3

2

1

y

0

-1

-2

-3

-4

-3

-2

-1

0

1

2

3

x

Figure 6.2.1

y = cos x and y = x

Unfortunately there is no trigonometric identity or simple method which will help us here.

Instead, we have to resort to numerical methods, which provide ways of getting successively

better approximations to the actual solution(s) to within any desired degree of accuracy.

There is a large field of mathematics devoted to this subject called numerical analysis. Many

of the methods require calculus, but luckily there is a method which we can use that requires

just basic algebra. It is called the secant method, and it finds roots of a given function f ( x),

i.e. values of x such that f ( x) = 0. A derivation of the secant method is beyond the scope of

this book,1 but we can state the algorithm it uses to solve f ( x) = 0:

1For an explanation of why the secant method works, see pp. 338-344 in A. RALSTON AND P. RABINOWITZ, A

First Course in Numerical Analysis, 2nd ed., New York: McGraw-Hill Book Co., 1978.

134

Chapter 6 • Additional Topics

§6.2

1. Pick initial points x 0 and x 1 such that x 0 < x 1 and f ( x 0) f ( x 1) < 0 (i.e. the solution is somewhere between x 0 and x 1).

2. For n ≥ 2, define the number xn by

( xn

x

−1 − xn−2) f ( xn−1)

n = xn−1 −

(6.2)

f ( xn−1) − f ( xn−2)

as long as | xn−1 − xn−2| > ǫerror, where ǫerror > 0 is the maximum amount of error desired

(usually a very small number).

3. The numbers x 0, x 1, x 2, x 3, ... will approach the solution x as we go through more iterations, getting as close as desired.

We will now show how to use this algorithm to solve the equation cos x = x. The solution

to that equation is the root of the function f ( x) = cos x − x. And we saw that the solution is

somewhere in the interval [0, 1]. So pick x 0 = 0 and x 1 = 1. Then f (0) = 1 and f (1) = −0.4597,

so that f ( x 0) f ( x 1) < 0 (we are using radians, of course). Then by definition,

( x 1 − x 0) f ( x 1)

x 2 = x 1 − f ( x 1) − f ( x 0)

(1 − 0) f (1)

= 1 − f (1) − f (0)

(1 − 0)(−0.4597)

= 1 −

−0.4597 − 1

= 0.6851 ,

( x 2 − x 1) f ( x 2)

x 3 = x 2 − f ( x 2) − f ( x 1)

(0.6851 − 1) f (0.6851)

= 0.6851 −

f (0.6851) − f (1)

(0.6851 − 1)(0.0893)

= 0.6851 − 0.0893 − (−0.4597)

= 0.7363 ,

and so on. Using a calculator is not very efficient and will lead to rounding errors. A bet-

ter way to implement the algorithm is with a computer. Listing 6.1 below shows the code

(secant.java) for solving cos x = x with the secant method, using the Java programming lan-

guage:

Numerical Methods in Trigonometry • Section 6.2

135

Listing 6.1

Program listing for secant.java

1

import java.math.*;

2

public class secant {

3

public static void main (String[] args) {

4

double x0 =

Double.parseDouble(args[0]);

5

double x1 =

Double.parseDouble(args[1]);

6

double x = 0;

7

double error = 1.0E-50;

8

for (int i=2; i <= 10; i++) {

9

if (Double.compare(Math.abs(x0 - x1),error) > 0) {

10

x = x1 - (x1 - x0)*f(x1)/(f(x1) - f(x0));

11

x0 = x1;

12

x1 = x;

13

System.out.println("x" + i + " = " + x);

14

} else {

15

break;

16

}

17

}

18

MathContext mc = new MathContext(50);

19

BigDecimal answer = new BigDecimal(x,mc);

20

System.out.println("x = " + answer);

21

}

22

//Define the function f(x)

23

public static double f (double x) {

24

return Math.cos(x) - x;

25

}

26

}

Lines 4-5 read in x 0 and x 1 as input parameters to the program.

Line 6 initializes the variable that will eventually hold the solution.

Line 7 sets the maximum error ǫerror to be 1.0 × 10−50. That is, our final answer will be

within that (tiny!) amount of the real solution.

Line 8 starts a loop of 9 iterations of the algorithm, i.e. it will create the successive ap-

proximations x 2, x 3, ..., x 10 to the real solution, though in Line 9 we check to see if the two

previous approximations differ by less than the maximum error. If they do, we stop (since

this means we have an acceptable solution), otherwise we continue.

Line 10 is the main step in the algorithm, creating xn from xn−1 and xn−2.

Lines 11-12 set the new values of xn−2 and xn−1, respectively.

Lines 18-20 set the number of decimal places to show in the final answer to 50 (the default

is 16) and then print the answer.

Lines 23-24 give the definition of the function f ( x) = cos x − x.

136

Chapter 6 • Additional Topics

§6.2

Below is the result of compiling and running the program using x 0 = 0 and x 1 = 1:

javac secant.java

java secant 0 1

x2 = 0.6850733573260451

x3 = 0.736298997613654

x4 = 0.7391193619116293

x5 = 0.7390851121274639

x6 = 0.7390851332150012

x7 = 0.7390851332151607

x8 = 0.7390851332151607

x = 0.73908513321516067229310920083662495017051696777344

Notice that the program only got up to x 8, not x 10. The reason is that the difference

between x 8 and x 7 was small enough (less than ǫerror = 1.0 × 10−50) to stop at x 8 and call

that our solution. The last line shows that solution to 50 decimal places.

Does that number look familiar? It should, since it is the answer to Exercise 11 in Section

4.1. That is, when taking repeated cosines starting with any number (in radians), you even-

tually start getting the above number repeatedly after enough iterations. This turns out not

to be a coincidence. Figure 6.2.2 gives an idea of why.

1

0.8

y = cos( x)

0.6

y

0.4

y = x

0.2

0

− π

-1

0

1

π

2

2

x

Figure 6.2.2

Attractive fixed point for cos x

Since x = 0.73908513321516... is the solution of cos x = x, you would get cos (cos x) =

cos x = x, so cos (cos (cos x)) = cos x = x, and so on. This number x is called an attractive fixed Numerical Methods in Trigonometry • Section 6.2

137

point of the function cos x. No matter where you start, you end up getting “drawn” to it.

Figure 6.2.2 shows what happens when starting at x = 0: taking the cosine of 0 takes you to

1, and then successive cosines (indicated by the intersections of the vertical lines with the

cosine curve) eventually “spiral” in a rectangular fashion to the fixed point (i.e. the solution),

which is the intersection of y = cos x and y = x.

Recall in Example 5.10 in Section 5.2 that we claimed that the maximum and minimum of

the function y = cos 6 x + sin 4 x were ±1.90596111871578, respectively. We can show this by

using the open-source program Octave.2 Octave uses a successive quadratic programming

method to find the minimum of a function f ( x). Finding the maximum of f ( x) is the same as

finding the minimum of − f ( x) then multiplying by −1 (why?). Below we show the commands

to run at the Octave command prompt (octave:n>) to find the minimum of f ( x) = cos 6 x +

sin 4 x. The command sqp(3,’f’) says to use x = 3 as a first approximation of the number x

where f ( x) is a minimum.

octave:1> format long

octave:2> function y = f(x)

> y = cos(6*x) + sin(4*x)

> endfunction

octave:3> sqp(3,’f’)

y = -1.90596111871578

ans = 2.65792064609274

The output says that the minimum occurs when x = 2.65792064609274 and that the min-

imum is −1.90596111871578. To find the maximum of f ( x), we find the minimum of − f ( x)

and then take its negative. The command sqp(2,’f’) says to use x = 2 as a first approxima-

tion of the number x where f ( x) is a maximum.

octave:4> function y = f(x)

> y = -cos(6*x) - sin(4*x)

> endfunction

octave:5> sqp(2,’f’)

y = -1.90596111871578

ans = 2.05446832062993

The output says that the maximum occurs when x = 2.05446832062993 and that the max-

imum is −(−1.90596111871578) = 1.90596111871578.

Recall from Section 2.4 that Heron’s formula is adequate for “typical” triangles, but will

often have a problem when used in a calculator with, say, a triangle with two sides whose

sum is barely larger than the third side. However, you can get around this problem by

using computer software capable of handling numbers with a high degree of precision. Most

2Freely available at http://www.gnu.org/software/octave

138

Chapter 6 • Additional Topics

§6.2

modern computer programming languages have this capability. For example, in the Python

programming language3 (chosen here for simplicity) the decimal module can be used to set

any level of precision.4 Below we show how to get accuracy up to 50 decimal places using

Heron’s formula for the triangle in Example 2.16 from Section 2.4, by using the python

interactive command shell:

>>> from decimal import *

>>> getcontext().prec = 50

>>> a = Decimal("1000000")

>>> b = Decimal("999999.9999979")

>>> c = Decimal("0.0000029")

>>> s = (a+b+c)/2

>>> K = s*(s-a)*(s-b)*(s-c)

>>> print Decimal(K).sqrt()

0.99999999999894999999999894874999999889618749999829

(Note: The triple arrow >>> is just a command prompt, not part of the code.)

Notice in this case that we do get the correct answer; the high level of precision eliminates

the rounding errors shown by many calculators when using Heron’s formula.

Another software option is Sage5, a powerful and free open-source mathematics package

based on Python. It can be run on your own computer, but it can also be run through a web

interface: go to http://sagenb.org to create a free account, then once you register and sign

in, click the New Worksheet link to start entering commands. For example, to find the

solution to cos x = x in the interval [0,1], enter these commands in the worksheet textfield:

x = var(’x’)

find_root(cos(x) == x, 0,1)

Click the evaluate link to display the answer: 0.7390851332151559

Exercises

1. One obvious solution to the equation 2 sin x = x is x = 0. Write a program to find the other

solution(s), accurate to at least within 1.0 × 10−20. You can use any programming language,

though you may find it easier to just modify the code in Listing 6.1 (only one line needs to be

changed!). It may help to use Gnuplot to get an idea of where the graphs of y = 2 sin x and y = x

intersect.

2. Repeat Exercise 1 for the equation sin x = x 2.

3. Use Octave or some other program to find the maximum and minimum of y = cos 5 x − sin 3 x.

3Available for free at http://www.python.org

4Other languages have similar capability, e.g. the BigDecimal class in Java.

5Visit the homepage at http://www.sagemath.org for more details.

Complex Numbers • Section 6.3

139

6.3 Complex Numbers

There is no real number x such that x 2 = −1. However, it turns out to be useful6 to invent

such a number, called the imaginary unit and denoted by the letter i. Thus, i 2 = −1, and

hence i = −1. If a and b are real numbers, then a number of the form a + bi is called a

complex number, and if b = 0 then it is called an imaginary number (and pure imagi-

nary if a = 0 and b = 0). The real number a is called the real part of the complex number

a + bi, and bi is called its imaginary part.

What does it mean to add a to bi in the definition a + bi of a complex number, i.e. adding

a real number and an imaginary number? You can think of it as a way of extending the set

of real numbers. If b = 0 then a + bi = a + 0 i = a (since 0 i is defined as 0), so that every real number is a complex number. The imaginary part bi in a + bi can be thought of as a way

of taking the one-dimensional set of all real numbers and extending it to a two-dimensional

set: there is a natural correspondence between a complex number a + bi and a point ( a, b) in

the (two-dimensional) x y-coordinate plane.

Before exploring that correspondence further, we will first state some fundamental prop-

erties of and operations on complex numbers:

Let a + bi and c + di be complex numbers. Then:

1. a + bi = c + di if and only if a = c and b = d (i.e. the real parts are equal and the imaginary parts are equal)

2. ( a + bi) + ( c + di) = ( a + c) + ( b + d) i (i.e. add the real parts together and add the imaginary parts together)

3. ( a + bi) − ( c + di) = ( a − c) + ( b − d) i

4. ( a + bi)( c + di) = ( ac − bd) + ( ad + bc) i

5. ( a + bi)( a − bi) = a 2 + b 2

a + bi

( ac + bd) + ( bc − ad) i

6.

=

c + di

c 2 + d 2

The first three items above are just definitions of equality, addition, and subtraction of

complex numbers. The last three items can be derived by treating the multiplication and

division of complex numbers as you would normally treat factors of real numbers:

( a + bi)( c + di) = a ( c + di) + bi ( c + di)

= ac + adi + bci + bdi 2 = ac + adi + bci + bd(−1)

= ( ac − bd) + ( ad + bc) i

6Especially in electrical engineering, physics, and various fields of mathematics.

140

Chapter 6 • Additional Topics

§6.3

The fifth item is a special case of the multiplication formula:

( a + bi)( a − bi) = (( a)( a) − ( b)(− b)) + (( a)(− b) + ( b)( a)) i

= ( a 2 + b 2) + (− ab + ba) i = ( a 2 + b 2) + 0 i

= a 2 + b 2

The sixth item comes from using the previous items:

a + bi

a + bi c − di

=

·

c + di

c + di c − di

( ac − b(− d)) + ( a(− d) + bc) i

=

c 2 + d 2

( ac + bd) + ( bc − ad) i

=

c 2 + d 2

The conjugate a + bi of a complex number a + bi is defined as a + bi = a − bi. Notice that ( a + bi) + ( a + bi) = 2 a is a real number, ( a + bi) − ( a + bi) = 2 bi is an imaginary number if b = 0, and ( a + bi)( a + bi) = a 2 + b 2 is a real number. So for a complex number z = a + bi, z z = a 2+ b 2 and thus we can define the modulus of z to be z z = a 2 + b 2, which we denote by | z|.

Example 6.9

Let z 1 = −2 + 3 i and z 2 = 3 + 4 i. Find z 1 + z 2, z 1 − z 2, z 1 z 2, z 1/ z 2, | z 1|, and | z 2|.

Solution: Using our rules and definitions, we have:

z 1 + z 2 = (−2 + 3 i) + (3 + 4 i)

= 1 + 7 i

z 1 − z 2 = (−2 + 3 i) − (3 + 4 i)

= −5 − i

z 1 z 2 = (−2 + 3 i)(3 + 4 i)

= ((−2)(3) − (3)(4)) + ((−2)(4) + (3)(3)) i

= −18 + i

z 1

−2 + 3 i

=

z 2

3 + 4 i

(−2)(3) + (3)(4) + ((3)(3) − (−2)(4)) i

=

32 + 42

6

17

=

+

i

25

25

| z 1| =

(−2)2 + 32

=

13

| z 2| =

32 + 42

= 5

Complex Numbers • Section 6.3

141

We know that any point ( x, y) in the x y-coordinate plane that is a distance r > 0 from the

origin has coordinates x = r cos θ and y = r sin θ, where θ is the angle in standard position as in Figure 6.3.1(a).

y

y

( x, y) = ( r cos θ, r sin θ)

z = x + yi = r cos θ + ( r sin θ) i

r

r

θ

x

θ

x

0

0

(a) Point ( x, y)

(b) Complex number z = x + yi

Figure 6.3.1

Let z = x + yi be a complex number. We can represent z as a point in the complex plane,

where the horizontal x-axis represents the real part of z, and the vertical y-axis represents

the pure imaginary part of z, as in Figure 6.3.1(b). The distance r from z to the origin is, by the Pythagorean Theorem, r =

x 2 + y 2, which is just the modulus of z. And we see from

Figure 6.3.1(b) that x = r cos θ and y = r sin θ, where θ is the angle formed by the positive x-axis and the line segment from the origin to z. We call this angle θ the argument of z.

Thus, we get the trigonometric form (sometimes called the polar form) of the complex

number z:

For any complex number z = x + yi, we can write

z = r (cos θ + i sin θ) , where

(6.3)

r = | z| =

x 2 + y 2 and

θ = the argument of z .

The representation z = r (cos θ + i sin θ) is often abbreviated as:

z = r cis θ

(6.4)

In the special case z = 0 = 0 + 0 i, the argument θ is undefined since r = | z| = 0. Also, note

that the argument θ can be replaced by θ + 360◦ k or θ + πk, depending on whether you are

using degrees or radians, respectively, for k = 0, ±1, ±2, ... . Note also that for z = x+ yi with

r = | z|, θ must satisfy

tan θ

y

y

=

, cos θ

, sin θ

.

x

= xr

= r

142

Chapter 6 • Additional Topics

§6.3

Example 6.10

Represent the complex number

y

−2 − i in trigonometric form.

Solution: Let z = −2 − i = x + yi, so that x = −2 and y = −1. Then θ is

θ

y

2

in QIII, as we see in Figure 6.3.2. So since tan θ

, we have

x

= x = −1

−2 = 1

2

0

θ = 206.6◦. Also,

1

r

r

z

=

x 2 + y 2 =

(−2)2 + (−1)2 =

5 .

= −2 − i

Figure 6.3.2

Thus, −2 − i = 5 (cos 206.6◦ + i sin 206.6◦) , or 5 cis 206.6◦.

For complex numbers in trigonometric form, we have the following formulas for multipli-

cation and division:

Let z 1 = r 1 (cos θ 1 + i sin θ 1) and z 2 = r 2 (cos θ 2 + i sin θ 2) be complex numbers. Then z 1 z 2 = r 1 r 2 (cos ( θ 1 + θ 2) + i sin ( θ 1 + θ 2)) , and

(6.5)

z 1

r 1

=

(cos ( θ 1 − θ 2) + i sin ( θ 1 − θ 2)) if z 2 = 0.

(6.6)

z 2

r 2

The proofs of these formulas are straightforward:

z 1 z 2 = r 1 (cos θ 1 + i sin θ 1) · r 2 (cos θ 2 + i sin θ 2)

= r 1 r 2 [(cos θ 1 cos θ 2 − sin θ 1 sin θ 2) + i (sin θ 1 cos θ 2 + cos θ 1 sin θ 2)]

= r 1 r 2 (cos ( θ 1 + θ 2) + i sin ( θ 1 + θ 2))

by the addition formulas for sine and cosine. And

z 1

r 1 (cos θ 1 + i sin θ 1)

=

z 2

r 2 (cos θ 2 + i sin θ 2)

r 1

cos θ 1 + i sin θ 1

cos θ 2 − i sin θ 2

=

·

·

r 2

cos θ 2 + i sin θ 2

cos θ 2 − i sin θ 2

r 1

(cos θ 1 cos θ 2 + sin θ 1 sin θ 2) + i (sin θ 1 cos θ 2 − cos θ 1 sin θ 2)

=

·

r 2

cos2 θ 2 + sin2 θ 2

r 1

=

(cos ( θ 1 − θ 2) + i sin ( θ 1 − θ 2))

r 2

by the subtraction formulas for sine and cosine, and since cos2 θ 2 + sin2 θ 2 = 1.

QED

Note that formulas (6.5) and (6.6) say that when multiplying complex numbers the moduli are multiplied and the arguments are added, while when dividing complex numbers the

moduli are divided and the arguments are subtracted. This makes working with complex

numbers in trigonometric form fairly simple.

Complex Numbers • Section 6.3

143

Example 6.11

Let z 1 = 6(cos 70◦ + i sin 70◦) and z 1 = 2(cos 31◦ + i sin 31◦). Find z 1 z 2 and z 1 .

z 2

Solution: By formulas (6.5) and (6.6) we have

z 1 z 2 = (6)(2)(cos (70◦ + 31◦) + i sin (70◦ + 31◦))

⇒

z 1 z 2 = 12(cos 101◦ + i sin 101◦) , and

z 1

6

z 1

=

(cos (70◦ − 31◦) + i sin (70◦ − 31◦))

⇒

= 3(cos 39◦ + i sin 39◦) .

z 2

2

z 2

For the special case when z 1 = z 2 = z = r (cos θ + i sin θ) in formula (6.5), we have

[ r (cos θ + i sin θ)]2 = r · r (cos ( θ + θ) + i sin ( θ + θ))

= r 2 (cos 2 θ + i sin 2 θ) ,

and so

[ r (cos θ + i sin θ)]3 = [ r (cos θ + i sin θ)]2 · r (cos θ + i sin θ)

= r 2 (cos 2 θ + i sin 2 θ) · r (cos θ + i sin θ)

= r 3 (cos (2 θ + θ) + i sin (2 θ + θ))

= r 3 (cos 3 θ + i sin 3 θ) ,

and continuing like this (i.e. by mathematical induction), we get:

Theorem 6.1. De Moivre’s Theorem:7 For any integer n ≥ 1,

[ r (cos θ + i sin θ)] n = rn (cos nθ + i sin nθ) .

(6.7)

We define z 0 = 1 and z− n = 1/ zn for all integers n ≥ 1. So by De Moivre’s Theorem and

formula (6.5), for any z = r (cos θ + i sin θ) and integer n ≥ 1 we get 1

z− n = zn

1 (cos 0◦ + i sin 0◦)

= rn(cos nθ + i sin nθ)

1

=

(cos (0◦ − nθ) + i sin (0◦ − nθ))

rn

= r− n (cos (− nθ) + i sin (− nθ)) ,

and so De Moivre’s Theorem in fact holds for all integers.8

7Named after the French statistician and mathematician Abraham de Moivre (1667-1754).

8There is a way of defining zn when n is a real (or complex) number, so that De Moivre’s Theorem holds for

any real number n. See pp. 59-60 in R.V. CHURCHILL, Complex Variables and Applications, 2nd ed., New York:

McGraw-Hill Book Co., 1960.

144

Chapter 6 • Additional Topics

§6.3

Example 6.12

Find (1 + i)10.

Solution: Since 1 + i = 2 (cos 45◦ + i sin 45◦) (why?), by De Moivre’s Theorem we have

(1 + i)10 = ( 2)10 (cos 450◦ + i sin 450◦) = 210/2 (0 + i (1)) = 25 · i = 32 i .

We can use De Moivre’s Theorem to find the nth roots of a complex number. That is, given

any complex number z and positive integer n, find all complex numbers w such that wn = z.

Let z = r (cos θ + i sin θ). Since the cosine and sine functions repeat every 360◦, we know that

z = r (cos ( θ +360◦ k) + i sin ( θ +360◦ k)) for k = 0, ±1, ±2, .... Now let w = r 0 (cos θ 0 + i sin θ 0) be an nth root of z. Then

wn = z

⇒

[ r 0 (cos θ 0 + i sin θ 0)] n = r (cos ( θ + 360◦ k) + i sin ( θ + 360◦ k))

⇒

rn (cos n

0

θ 0 + i sin nθ 0) = r (cos ( θ + 360◦ k) + i sin ( θ + 360◦ k))

⇒

rn 0 = r and nθ 0 = θ + 360◦ k

θ + 360◦ k

⇒

r 0 = r 1/ n and θ 0 =

.

n

Since the cosine and sine of θ+360◦ k will repeat for k

n

≥ n, we get the following formula for the

nth roots of z:

For any nonzero complex number z = r (cos θ + i sin θ) and positive integer n, the n

distinct nth roots of z are

θ + 360◦ k

θ + 360◦ k

r 1/ n cos

+ i sin

(6.8)

n

n

for k = 0, 1, 2, ..., n − 1.

Note: An nth root of z is usually written as z 1/ n or n z. The number r 1/ n in the above formula is the usual real nth root of the real number r = | z|.

Example 6.13

Find the three cube roots of i.

Solution: Since i = 1(cos 90◦ + i sin 90◦), the three cube roots of i are:

3

90◦ + 360◦(0)

90◦ + 360◦(0)

3

1

1 cos

+ i sin

= cos 30◦ + i sin 30◦

=

+

i ,

3

3

2

2

3

90◦ + 360◦(1)

90◦ + 360◦(1)

3

1

1 cos

+ i sin

= cos 150◦ + i sin 150◦ = −

+

i ,

3

3

2

2

3

90◦ + 360◦(2)

90◦ + 360◦(2)

1 cos

+ i sin

= cos 270◦ + i sin 270◦ = − i

3

3

Complex Numbers • Section 6.3

145

Notice from Example 6.13 that the three cube roots of i

y

are equally spaced points along the unit circle

| z| = 1

| z| = 1 in the

complex plane, as shown in Figure 6.3.3. We see that consec-

120◦

3

− 3

utive cube roots are 120◦ apart. In general, the n nth roots of

2 + i 2

2 + i 2

x

a complex number z will be equally spaced points along the

120◦

120◦

circle of radius | z|1/ n in the complex plane, with consecutive

roots separated by 360◦ .

n

In higher mathematics the Fundamental Theorem of Alge-

− i

bra states that every polynomial of degree n with complex

Figure 6.3.3

coefficients has n complex roots (some of which may repeat).

In particular, every real number a has n nth roots (being the roots of zn − a). For example,

the square roots of 1 are ±1, and the square roots of −1 are ± i.

Exercises

For Exercises 1-16, calculate the given expression.

1. (2 + 3 i) + (−3 − 2 i)

2. (2 + 3 i) − (−3 − 2 i)

3. (2 + 3 i) · (−3 − 2 i)

4. (2 + 3 i)/(−3 − 2 i)

5. (2 + 3 i) + (−3 − 2 i)

6. (2 + 3 i) − (−3 − 2 i)

7. (1 + i)/(1 − i)

8. |−3 + 2 i|

9. i 3

10. i 4

11. i 5

12. i 6

13. i 7

14. i 8

15. i 9

16. i 2009

For Exercises 17-24, prove the given identity for all complex numbers.

17. ( z) = z

18. z 1 + z 2 = z 1 + z 2

19. z 1 − z 2 = z 1 − z 2

20. z 1 z 2 = z 1 z 2

z

z

| z

1

z 1

1

1 |

21.

=

22. | z| = | z|

23. | z 1 z 2| = | z 1|| z 2|

24.

=

z

z

| z

2

z 2

2

2 |

For Exercises 25-30, put the given number in trigonometric form.

25. 2 + 3 i

26. −3 − 2 i

27. 1 − i

28. − i

29. 1

30. −1

31. Verify that De Moivre’s Theorem holds for the power n = 0.

For Exercises 32-35, calculate the given number.

32. 3 (cos 14◦ + i sin 14◦) · 2(cos 121◦ + i sin 121◦)

3 (cos 14◦ + i sin 14◦)

33. [3 (cos 14◦ + i sin 14◦)]4

34. [3 (cos 14◦ + i sin 14◦)]−4

35. 2(cos 121◦ + i sin 121◦)

36. Find the three cube roots of − i.

37. Find the three cube roots of 1 + i.

38. Find the three cube roots of 1.

39. Find the three cube roots of −1.

40. Find the five fifth roots of 1.

41. Find the five fifth roots of −1.

42. Find the two square roots of −2 + 2 3 i.

43. Prove that if z is an nth root of a real number a, then so is z. ( Hint: Use Exercise 20. ) 146

Chapter 6 • Additional Topics

§6.4

6.4 Polar Coordinates

Suppose that from the point (1, 0) in the

y

x y-coordinate plane we draw a spiral around

the origin, such that the distance between

any two points separated by 360◦ along the

← 1

spiral is always 1, as in Figure 6.4.1. We

←

→ 1

can not express this spiral as y = f ( x) for

→

some function f in Cartesian coordinates,

since its graph violates the vertical rule.

x

0

1

2

3

However, this spiral would be simple to

describe using the polar coordinate system.

Recall that any point P distinct from the

origin (denoted by O) in the x y-coordinate

plane is a distance r > 0 from the origin,

−−→

and the ray OP makes an angle θ with the

positive x-axis, as in Figure 6.4.2. We call

the pair ( r, θ) the polar coordinates of P,

Figure 6.4.1

and the positive x-axis is called the polar

axis of this coordinate system. Note that ( r, θ) = ( r, θ +360◦ k) for k = 0, ±1, ±2, ..., so (unlike for Cartesian coordinates) the polar coordinates of a point are not unique.

y

y

P( r, θ)

r

θ

θ

x

x

O

O

− r

P(− r, θ)

Figure 6.4.2

Polar coordinates ( r, θ)

Figure 6.4.3

Negative r: (− r, θ)

In polar coordinates we adopt the convention that r can be negative, by defining (− r, θ) =

−−→

( r, θ + 180◦) for any angle θ. That is, the ray OP is drawn in the opposite direction from the

angle θ, as in Figure 6.4.3. When r = 0, the point ( r, θ) = (0, θ) is the origin O, regardless of the value of θ.

You may be familiar with graphing paper, for plotting points or functions given in Carte-

sian coordinates (sometimes also called rectangular coordinates). Such paper consists of a

rectangular grid. Similar graphing paper exists for plotting points and functions in polar

coordinates, similar to Figure 6.4.4.

Polar Coordinates • Section 6.4

147

90◦

105◦

75◦

120◦

60◦

135◦

45◦

150◦

30◦

165◦

15◦

180◦

O

0◦

195◦

345◦

210◦

330◦

225◦

315◦

240◦

300◦

255◦

285◦

270◦

Figure 6.4.4

Polar coordinate graph

The angle θ can be given in either degrees or radians, whichever is more convenient.

Radians are often preferred when graphing functions in polar coordinates. The reason is

that, unlike degrees, radians can be considered “unitless” (as we mentioned in Chapter 4).

This is desirable when a function given in polar coordinates is expressed as r as a function

of θ (similar to how, in Cartesian coordinates ( x, y), functions are usually expressed as y as

a function of x). For example, if a function in polar coordinates is written as r = 2 θ, then r

would have the same units as θ. But r should be a unitless quantity, hence using radians for

θ makes more sense in this case.

148

Chapter 6 • Additional Topics

§6.4

Example 6.14

Express the spiral from Figure 6.4.1 in polar coordinates.

Solution: We will use radians for θ. The goal is to find some equation involving r and θ that describes

the spiral. We see that

θ = 0

⇒

r = 1

θ = 2 π

⇒

r = 2

θ = 4 π

⇒

r = 3

...

θ = 2 π k

⇒

r = 1 + k

for k = 0,1,2,.... In fact, that last relation holds for any nonnegative real number k (why?). So for

any θ ≥ 0,

θ

θ

θ = 2 π k

⇒

k =

⇒

r = 1 + k = 1 +

.

2 π

2 π

θ

Hence, the spiral can be written as r = 1 +

for θ ≥ 0. The graph is shown in Figure 6.4.5, along

2 π

with the Gnuplot commands to create the graph.

4

set polar

set size square

3

set samples 2000

unset key

2

set zeroaxis

set xlabel "x"

1

set ylabel "y"

plot [0:6*pi] 1 + t/(2*pi)

y

0

1

2

3

4

4

3

2

1

0

1

2

3

4

x

Figure 6.4.5

r = 1 + θ

2 π

Note that when using the set polar command, Gnuplot will assume that the function being plotted

is r as a function of θ (represented by the variable t in Gnuplot).

Polar Coordinates • Section 6.4

149

y

Figure 6.4.6 shows how to convert between polar coordinates and

( x, y) ( r, θ)

Cartesian coordinates. For a point with polar coordinates ( r, θ) and

Cartesian coordinates ( x, y):

r

y

Polar to Cartesian:

θ

x

x

O

= r cos θ

y = r sin θ

(6.9)

x

Cartesian to Polar:

Figure 6.4.6

y

r = ±

x 2 + y 2

tan θ =

if x = 0

(6.10)

x

Note that in formula (6.10), if x = 0 then θ = π/2 or θ = 3 π/2. Also, if x = 0 and y = 0 then the two possible solutions for θ in the equation tan θ

y

=

are in opposite quadrants (for

x

0 ≤ θ < 2 π). If the angle θ is in the same quadrant as the point ( x, y), then r =

x 2 + y 2 (i.e.

r is positive); otherwise r = − x 2 + y 2 (i.e. r is negative).

Example 6.15

Convert the following points from polar coordinates to Cartesian coordinates:

(a) (2, 30◦); (b) (3, 3 π/4); (c) (−1,5 π/3)

Solution: (a) Using formula (6.9) with r = 2 and θ = 30◦, we get:

( x, y) = ( r cos θ, r sin θ) = (2 cos 30◦,2 sin 30◦) = 2 · 3 ,2

2

· 12

⇒

( x, y) =

3, 1

(b) Using formula (6.9) with r = 3 and θ = 3 π/4, we get:

( x, y) = ( r cos θ, r sin θ) = 3 cos 3 π ,3 sin 3 π

, 3

, 3

4

4

= 3 · −1

· 1

⇒

( x, y) = −3

2

2

2

2

(c) Using formula (6.9) with r = −1 and θ = 5 π/3, we get:

( x, y) = ( r cos θ, r sin θ) = −1 cos 5 π ,

,

, 3

3

−1 sin 5 π

3

= −1 · 12 −1 · − 3

2

⇒

( x, y) = − 12 2

Example 6.16

Convert the following points from Cartesian coordinates to polar coordinates:

(a) (3, 4); (b) (−5,−5)

Solution: (a) Using formula (6.10) with x = 3 and y = 4, we get:

y

4

tan θ =

=

⇒

θ = 53.13◦ or θ = 233.13◦

x

3

Since θ = 53.13◦ is in the same quadrant (QI) as the point ( x, y) = (3,4), we can take

r =

x 2 + y 2 = 32 + 42 = 5. Thus, ( r, θ) = (5,53.13◦) .

Note that if we had used θ = 233.13◦, then we would have ( r, θ) = (−5,233.13◦).

150

Chapter 6 • Additional Topics

§6.4

(b) Using formula (6.10) with x = −5 and y = −5, we get:

y

−5

tan θ =

=

= 1

⇒

θ = 45◦ or θ = 225◦

x

−5

Since θ = 225◦ is in the same quadrant (QIII) as the point ( x, y) = (−5,−5), we can take

r =

x 2 + y 2 =

(−5)2 + (−5)2 = 5 2. Thus, ( r, θ) = (5 2,225◦) .

Note that if we had used θ = 45◦, then we would have ( r, θ) = (−5 2,45◦).

Example 6.17

Write the equation x 2 + y 2 = 9 in polar coordinates.

Solution: This is just the equation of a circle of radius 3 centered at the origin. Since r = ± x 2 + y 2 =

± 9, in polar coordinates the equation can be written as simply r = 3 .

Example 6.18

Write the equation x 2 + ( y − 4)2 = 16 in polar coordinates.

Solution: This is the equation of a circle of radius 4 centered at the point (0, 4). Expanding the

equation, we get:

x 2 + ( y − 4)2 = 16

x 2 + y 2 − 8 y + 16 = 16

x 2 + y 2 = 8 y

r 2 = 8 r sin θ

r = 8 sin θ

Why could we cancel r from both sides in the last step? Because we know that the point (0, 0) is on

the circle, so canceling r does not eliminate r = 0 as a potential solution of the equation (since θ = 0◦

would make r = 8 sin θ = 8 sin 0◦ = 0). Thus, the equation is r = 8 sin θ .

Example 6.19

Write the equation y = x in polar coordinates.

Solution: This is the equation of a line through the origin. So when x = 0, we know that y = 0. When

x = 0, we get:

y = x

y = 1

x

tan θ = 1

θ = 45◦

Since there is no restriction on r, we could have r = 0 and θ = 45◦, which would take care of the case

x = 0 (since then ( x, y) = (0,0), which is the same as ( r, θ) = (0,45◦)). Thus, the equation is θ = 45◦ .

Polar Coordinates • Section 6.4

151

Example 6.20

Prove that the distance d between two points ( r 1, θ 1) and ( r 2, θ 2) in polar coordinates is

d =

r 21 + r 22 − 2 r 1 r 2 cos ( θ 1 − θ 2) .

(6.11)

Solution: The idea here is to use the distance formula in Cartesian coordinates, then convert that

to polar coordinates. So write

x 1 = r 1 cos θ 1

y 1 = r 1 sin θ 1

x 2 = r 2 cos θ 2

y 2 = r 2 sin θ 2 .

Then ( x 1, y 1) and ( x 2, y 2) are the Cartesian equivalents of ( r 1, θ 1) and ( r 2, θ 2), respectively. Thus, by the Cartesian coordinate distance formula,

d 2 = ( x 1 − x 2)2 + ( y 1 − y 2)2

= ( r 1 cos θ 1 − r 2 cos θ 2)2 + ( r 1 sin θ 1 − r 2 sin θ 2)2

= r 2 cos2

cos2

sin2

sin2

1

θ 1 − 2 r 1 r 2 cos θ 1 cos θ 2 + r 22

θ 2 + r 21

θ 1 − 2 r 1 r 2 sin θ 1 sin θ 2 + r 22

θ 2

= r 2(cos2

(cos2

1

θ 1 + sin2 θ 1) + r 22

θ 2 + sin2 θ 2) − 2 r 1 r 2(cos θ 1 cos θ 2 + sin θ 1 sin θ 2)

d 2 = r 21 + r 22 − 2 r 1 r 2 cos ( θ 1 − θ 2) ,

so the result follows by taking square roots of both sides.

In Example 6.17 we saw that the equation x 2 + y 2 = 9 in Cartesian coordinates could be

expressed as r = 3 in polar coordinates. This equation describes a circle centered at the

origin, so the circle is symmetric about the origin. In general, polar coordinates are useful

in situations when there is symmetry about the origin (though there are other situations),

which arise in many physical applications.

Exercises

For Exercises 1-5, convert the given point from polar coordinates to Cartesian coordinates.

1. (6, 210◦)

2. (−4,3 π)

3. (2, 11 π/6)

4. (6, 90◦)

5. (−1,405◦)

For Exercises 6-10, convert the given point from Cartesian coordinates to polar coordinates.

6. (3, 1)

7. (−1,−3)

8. (0, 2)

9. (4, −2)

10. (−2,0)

For Exercises 11-18, write the given equation in polar coordinates.

11. ( x − 3)2 + y 2 = 9

12. y = − x

13. x 2 − y 2 = 1

14. 3 x 2 + 4 y 2 − 6 x = 9

15. Graph the function r = 1 + 2 cos θ in polar coordinates.

Appendix A

Answers and Hints to Selected Exercises

Chapter 1

13. sin A = 6/ 10, tan A = 6/2,

csc A = 10/ 6, sec A = 10/2, cot A = 2/ 6

Section 1.1 (p. 5)

15. sin A = 5/ 106, cos A = 9/ 106,

csc A = 106/5, sec A = 106/9, cot A = 9/5

1. 115◦

3. A = 52◦, B = 104◦

5. 45◦

17. sin A = 40/7, cos A = 3/7,

7. A = 9◦, B = 81◦

8. 0.011◦ and 89.989◦

tan A = 40/3, csc A = 7/ 40, cot A = 3/ 40

9. 25 miles

10. 111.8 ft

19. cos 3◦

21. sin 44◦

23. csc 13◦

15. Hint: Are the opposite sides of the four-

25. sin 77◦

27. tan 80◦

30. Hint: Draw a

sided figure inside the circle parallel?

right triangle with an acute angle A.

33. Hint: Draw two right triangles whose hy-

Section 1.2 (p. 12)

potenuses are the same length.

37. (a)

13/4 (b) 4 3/ 13 (c) 3/ 13

1. sin A = 5/13, cos A = 12/13, tan A = 5/12,

csc A = 13/5, sec A = 13/12, cot A = 12/5;

Section 1.3 (p. 20)

sin B = 12/13, cos B = 5/13, tan B = 12/5,

csc B = 13/12, sec B = 13/5, cot B = 5/12

1. 102.7 ft

3. 241.1 ft

4. 274 ft

7. 1062

3. sin A = 7/25, cos A = 24/25, tan A = 7/24,

mi

9. 0.476 in

11. 1.955 in

13. 0.4866

csc A = 25/7, sec A = 25/24, cot A = 24/7;

in

14. Partial answer: DE = a cot θ cos2 θ

sin B = 24/25, cos B = 7/25, tan B = 24/7,

15. c = 13, A = 22.6◦, B = 67.4◦

17. a =

csc B = 25/24, sec B = 25/7, cot B = 7/24

0.28, c = 2.02, B = 82◦

19. b = 6.15, c =

5. sin A = 9/41, cos A = 40/41, tan A = 9/40,

6.84, B = 64◦

csc A = 41/9, sec A = 41/40, cot A = 40/9;

21. a = 6.15, c = 6.84, A = 64◦

23. a = 2,

sin B = 40/41, cos B = 9/41, tan B = 40/9,

b = 2, B = 45◦

25. (a) 0.944 cm

csc B = 41/40, sec B = 41/9, cot B = 9/40

(b) 2.112 cm

27. (a)

3 a

(b) 35.26◦

7. sin A = 1/ 10, cos A = 3/ 10, tan A = 1/3,

29. 1379.5 ft = 0.2613 mi

csc A = 10, sec A = 10/3, cot A = 3;

sin B = 3/ 10, cos B = 1/ 10, tan B = 3,

Section 1.4 (p. 31)

csc B = 10/3, sec B = 10, cot B = 1/3

9. sin A = 5/6, cos A = 11/6, tan A = 5/ 11,

1. QII

3. QIV

5. negative y-axis

csc A = 6/5, sec A = 6/ 11, cot A = 11/5;

7. QIII

9. QIV

11. QI, QIII

13. QI,

sin B = 11/6, cos B = 5/6, tan B = 11/5,

QIV

15. QI, QII

17. 43◦

19. 54◦

21.

csc B = 6/ 11, sec B = 6/5, cot B = 5/ 11

85◦

23.

sin θ =

3/2 and tan θ = − 3;

11. cos A = 7/4, tan A = 3/ 7, csc A = 4/3, sin θ = − 3/2 and tan θ = 3

sec A = 4/ 7, cot A = 7/3

25. sin θ = 21/5 and tan θ = 21/2;

152

Appendix A:

Answers and Hints to Selected Exercises

153

sin θ = − 21/5 and tan θ = − 21/2

Section 2.4 (p. 58)

27. cos θ = 3/2 and tan θ = 1/ 3;

cos θ = − 3/2 and tan θ = −1/ 3

1. 22.55

3. 9.21

5. 3 15

4

≈ 2.905

29. cos θ = ±1 and tan θ = 0

7. 12.21

9. Hints: The diagonals break the

31. cos θ = 0 and tan θ is undefined

quadrilateral into four triangles; also, con-

33. sin θ = 1/ 5 and cos θ = −2/ 5;

sider formulas (2.23)-(2.25).

sin θ = −1/ 5 and cos θ = 2/ 5

35. sin θ = 5/13 and cos θ = 12/13;

Section 2.5 (p. 64)

sin θ = −5/13 and cos θ = −12/13

37. No

39. No

1. R = 2.63, r = 0.69

3. R = 3.51, r = 1.36

5. R = 24.18, r = 1.12

12. (c) Twice as large

Section 1.5 (p. 37)

(d) Hint: Bisect each angle.

1. (a) 328◦ (b) 148◦ (c) 212◦

3. (a) 248◦

(b) 68◦ (c) 292◦

7. 25◦, 155◦

9. 65◦, 295◦

11. 38◦, 218◦

13. 169◦, 191◦

Chapter 3

15. D =

ab 2 , a 2 b

a 2+ b 2 a 2+ b 2

Section 3.1 (p. 70)

Chapter 2

1.

θ = 270◦

3.

Hint:

See Example

3.7.

19.

tan θ = ± sin θ/ 1 − sin2 θ =

Section 2.1 (p. 43)

± 1 − cos2 θ/cos θ

1. b = 7.4, c = 15.1, C = 120◦

3. a = 9.7,

b = 10.7, C = 95◦

5. b = 65.1, B = 136.5◦, Section 3.2 (p. 76)

C = 18.5◦

7. No solution

9. b = 24.9,

B = 59.9◦, C = 70.1◦; b = 9.9, B = 20.1◦, C = 3. sin ( A + B) = 1020, cos ( A

,

1189

+ B) = − 611

1189

109.9◦

11. 422 mi/hr

15. 5.66 cm and

tan ( A + B) = − 1020

4. ( 6

611

+ 2)/4

12.86 cm

16. Hint: Think geometrically.

5. 2 − 3

15. Hint: For a = 0 and b = 0,

draw a right triangle with legs of lengths a

Section 2.2 (p. 49)

and b.

1. a = 10.6, B = 40.9◦, C = 79.1

3. A =

47.9◦, b = 8.2, C = 72.1◦

5. No solution

7.

Section 3.3 (p. 81)

4.13 and 8.91 cm

9. 50.5◦, 59◦, 70.5◦

11. 7 cm

15. Hints: One of the angles in

9. Hint: Is sin A + cos A always positive? 11.

the formulas is a right angle; also, use the

1/2

definition of cosine.

Section 2.3 (p. 53)

Section 3.4 (p. 86)

1. A = 79.1◦, B = 40.9◦, c = 10.6

3. A = 13. Hint: One way to do this is with the Law

47.9◦, b = 8.2, C = 72.1◦

5. No

6. Yes

of Tangents. Another way is with the Law of

11. Hint: Think of Exercise 10.

Sines.

154

Appendix A:

Answers and Hints to Selected Exercises

Chapter 4

13. max. at x = ± π/2, ± 5 π/2, ± 9 π/2, ...

min. at x = ± 3 π/2, ± 7 π/2, ± 11 π/2, ...

Section 4.1 (p. 89)

15. amplitude = 0.5, period = π

17. out of

1.

phase

18. in phase

19. amplitude

π/45

3. 13 π/18

5. −3 π/5

7. 36◦

9.

= 34,

174◦

period = 2

21. amplitude = 2 2, period

= 2 π

23. 2 π

25. 6

27. amplitude en-

Section 4.2 (p. 94)

velope: y = ± x 2

29. No

1. 9.6 cm

3. 11 π in

5. 54.94 in

Section 5.3 (p. 128)

7. 12.86 ft

8. 34.18

9. 38.26

11. 3.392 and 9.174

12. 3.105828541

1. π/4

3. 0

5. π

7. π/2

9. 0

11. − π/3

13. π/7

15. 4 π/5

17. π/6

Section 4.3 (p. 98)

19. − π/9

21. 12/13

23. π/2

25. π/2

1. 1.512 cm2

3. 24.5 m2

5. 269.1 cm2

Chapter 6

7. 5 cm2

9. π/2 cm2

11. 0.017 cm2

13.

21.46

15. 48.17

17. 0.522 m2

Section 6.1 (p. 132)

19. Sector area is quadrupled, arc length is

doubled.

1. 3 π

5.

4 + πk

3. 3 π

10 + 2 πk

5

± π 6 + πk

7. −0.821 + 2 πk, 3.963 + 2 πk

9. π 4 + πk

Section 4.4 (p. 102)

11. 2 πk

3

1. ν = 6 m/sec, ω = 1.5 rad/sec

Section 6.2 (p. 138)

3. ν = 6.6 m/sec, ω = 0.94 rad/sec

5. ν = 3.75 m/sec, ω = 1.875 rad/sec

1. x = 1.89549426703398093962

7. 3.375 rad

9. 32 rpm and 21.33 rpm

11. 40.84 in/sec

Section 6.3 (p. 145)

1. −1 + i

3. −13 i

5. −1 − i

7. i

Chapter 5

9. − i

11. i

13. − i

15. i

17.

Let z = a + bi.

Then z = a − bi, so

Section 5.1 (p. 108)

( z) = a − bi = a+ bi = z.

23. Hint: Use Exer-

13. Partial answer: sec θ = OQ

cise 20.

25.

13 cis 56.3◦

27.

2 cis 315◦

29. cis 0◦

33. 81 cis 56◦

35. 1.5 cis 253◦

Section 5.2 (p. 118)

37. 6 2 cis 15◦, 6 2 cis 135◦, 6 2 cis 255◦

39. 1

i,

i

41. cis 36◦, cis 108◦,

2 +

3

2

−1, 12 − 3

2

1. amplitude = 3, period = 2, phase shift = 0 cis 180◦, cis 252◦, cis 324◦

3. amplitude = 1, period = 2 π/5, phase shift =

−3/5

5. amplitude = 1, period = 2 π/5, phase Section 6.4 (p. 151)

shift = − π/5

7. amplitude = 1, period = π,

phase shift = 3 π/2

1. (−3 3,−3)

3. ( 3, −1)

9. amplitude undefined, period = π/2, phase 5. (−1/ 2,−1/ 2)

7. ( 10, 251.6◦)

shift = 3 π/2

11. amplitude undefined, pe-

9. (2 5, 333.4◦)

11. r = 6 cos θ

riod = π, phase shift = 1/2

13. r 2 cos 2 θ = 1

14. r = 3/(2 − cos θ)

Appendix B

Graphing with Gnuplot

Gnuplot is a free, open-source software package for producing a variety of graphs. Versions

are available for many operating systems. Below is a very brief tutorial on how to use

Gnuplot to graph trigonometric functions.

INSTALLATION

1. Go to http://www.gnuplot.info/download.html and follow the links to download the lat-

est version for your operating system. For Windows, you should download the setup file

with a name such as gp460-win32-setup.exe, which is version 4.6.0. All the examples

discussed here will assume at least version 4.6.0, though they should work with earlier

4.x versions.

2. Install the downloaded file. For example, in Windows you would run the setup file you

downloaded in Step 1, which installs Gnuplot in the C:\Program Files\gnuplot folder by

default. You can accept the defaults during installation, though you should select the

“Create a desktop icon” option in the Select Additional Tasks screen.

3. (Optional) Read the documentation at http://gnuplot.info/documentation.html.

RUNNING GNUPLOT

1. In Windows, run wgnuplot.exe from the bin folder where you installed Gnuplot (the de-

fault location is C:\Program Files\gnuplot\bin\wgnuplot.exe), or double-click the desk-

top icon if you selected that option during the installation. In Linux, just type gnuplot in

a terminal window.

2. You should now get a Gnuplot terminal with a gnuplot> command prompt. In Windows

this will appear in a new window, as shown in the picture on the next page. In Linux it

will appear in the terminal window where the gnuplot command was run. For Windows,

if the font is unreadable you can change it by right-clicking on the text part of the Gnu-

plot window and selecting the “Choose Font..” option. For example, the font “Courier”,

style “Regular”, size “12” is usually a good choice (that choice can be saved for future ses-

sions by right-clicking in the Gnuplot window again and selecting the option to update

wgnuplot.ini).

3. At the gnuplot> command prompt you can now run graphing commands, which we will

now describe.

155

156

Appendix B:

Graphing with Gnuplot

GRAPHING FUNCTIONS

The usual way to create graphs in Gnuplot is with the plot command:

plot <range> <comma-separated list of functions>

For a function y = f ( x), <range> is the range of x values (and optionally the range of y values) over which to plot. To specify an x range, use an expression of the form [ a : b], for

some numbers a < b. This will cause the graph to be plotted for a ≤ x ≤ b.

To specify an x range and a y range, use an expression of the form [ a : b][ c : d], for some numbers a < b and c < d. This will cause the graph to be plotted for a ≤ x ≤ b and c ≤ y ≤ d.

Function definitions use the x variable in combination with mathematical operators, listed

below:

Appendix B:

Graphing with Gnuplot

157

Symbol

Operation

Example

Result

+

Addition

2 + 3

5

−

Subtraction

3 − 2

1

*

Multiplication

2*3

6

/

Division

4/2

2

**

Power

2**3

23 = 8

exp( x)

ex

exp(2)

e 2

log( x)

ln x

log(2)

ln 2

sin( x)

sin x

sin(pi/2)

1

cos( x)

cos x

cos(pi)

−1

tan( x)

tan x

tan(pi/4)

1

Note that we use the special keyword “pi” to denote the value of π.

Example B.1. To graph the function y = sin x from x = 0 to x = 2 π, type this at the gnuplot> prompt:

plot [0:2*pi] sin(x)

The result is shown below:

1

sin( x)

0.8

0.6

0.4

0.2

0

-0.2

-0.4

-0.6

-0.8

-1

0

1

2

3

4

5

6

Notice that the x-axis is labeled with integers. To get the x-axis labels with fractions of π,

you need to modify the terminal setting. In Windows, you would do this:

set terminal windows enhanced

158

Appendix B:

Graphing with Gnuplot

In Linux you would do this:

set terminal wxt enhanced

You can then (provided the Symbol font is installed, which it usually is) set the x-axis to

have multiples of π/2 from 0 to 2 π as labels with this command (all on one line):

set xtics (’0’ 0,’{/Symbol p}/2’ pi/2,’{/Symbol p}’ pi,’3{/Symbol p}/2’ 3*pi/2,

’2{/Symbol p}’ 2*pi)

In the above example, to also plot the function y = cos 2 x + sin 3 x on the same graph, put a

comma after the first function then append the new function:

plot [0:2*pi] sin(x), cos(2*x) + sin(3*x)

By default, the x-axis is not shown in the graph. To display it, use this command before the

plot command:

set zeroaxis

Also, to label the axes, use these commands:

set xlabel "x"

set ylabel "y"

The default sample size for plots is 100 units, which can result in jagged edges if the curve

is complicated. To get a smoother curve, increase the sample size (to, say, 1000) like this:

set samples 1000

Putting all this together, we get the following graph:

Appendix B:

Graphing with Gnuplot

159

2

sin( x)

cos(2 ∗ x) + sin(3 ∗ x)

1.5

1

0.5

y

0

-0.5

-1

-1.5

-2

0

π/2

π

3 π/2

2 π

x

PRINTING AND SAVING

In Windows, if you are using the windows enhanced terminal then to print a graph from

Gnuplot click on the printer icon in the menubar of the graph’s window. If you are using the

default wxt terminal then select Print near the top of the main Gnuplot window and enter

png in the Terminal type? textfield, then hit OK to get the Print Setup dialog.

In Windows, to save a graph, say, as a PNG file, go to the File menu on the main Gnu-

plot menubar, select “Output Device ...”, and enter png in the Terminal type? textfield, hit

OK. Then, in the File menu again, select the “Output ...” option and enter a filename (say,

graph.png) in the Output filename? textfield, hit OK. Now run your plot command again and

the file will be saved in the current directory, usually in your My Documents folder (it can also

be found by selecting the “show Current Directory” option in the File menu).

In Linux, to save the graph as a file called graph.png run the following commands:

set terminal png

set output ’graph.png’

and then run your plot command. There are many terminal types (which determine the

output format). Run the command set terminal to see all the possible types. In Linux,

the postscript terminal type is popular, since the print quality is high and there are many

PostScript viewers available.

To quit Gnuplot, type quit at the gnuplot> command prompt.

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ADDENDUM: How to use this License for your documents

To use this License in a document you have written, include a copy of the License in the

document and put the following copyright and license notices just after the title page:

Copyright © YEAR YOUR NAME. Permission is granted to copy, distribute and/or

modify this document under the terms of the GNU Free Documentation License, Ver-

sion 1.3 or any later version published by the Free Software Foundation; with no

Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the

license is included in the section entitled “GNU Free Documentation License”.

If you have Invariant Sections, Front-Cover Texts and Back-Cover Texts, replace the “with

. . . Texts.” line with this:

with the Invariant Sections being LIST THEIR TITLES, with the Front-Cover Texts

being LIST, and with the Back-Cover Texts being LIST.

If you have Invariant Sections without Cover Texts, or some other combination of the

three, merge those two alternatives to suit the situation.

If your document contains nontrivial examples of program code, we recommend releas-

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General Public License, to permit their use in free software.

History

This section contains the revision history of the book. For persons making modifications to

the book, please record the pertinent information here, following the format in the first item

below.

1. VERSION: 1.0

Date: 2009-07-01

Author(s): Michael Corral

Title: Trigonometry

Modification(s): Initial version

2. VERSION: 1.1

Date: 2010-03-12

Author(s): Michael Corral

Title: Trigonometry

Modification(s): Made the following changes:

a) Added a section on polar coordinates

b) Changed the format of the page headers

c) Changed the format of the examples

3. VERSION: 1.2

Date: 2012-06-14

Author(s): Michael Corral

Title: Trigonometry

Modification(s): Re-licensed under the terms of the GNU Free Documentation License

version 1.3.

168

Index

Symbols

arc cosecant . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

⊥. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4

arc cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

∼ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

arc cotangent. . . . . . . . . . . . . . . . . . . . . . . . . . 126

arc length . . . . . . . . . . . . . . . . . . . . . . . . . . . 90, 91

A

arc secant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

acute angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

arc sine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

acute triangle . . . . . . . . . . . . . . . . . . . . . . . . . . 38

arc tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

addition formulas . . . . . . . . . . . . . . . . . . . . . . 71

area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Ahmes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

altitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

ambiguous case in Law of Sines . . . . . . . . 41

triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

argument of a complex number . . . . . . . 141

amplitude envelope . . . . . . . . . . . . . . . . . . . 119

asymptote . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

horizontal . . . . . . . . . . . . . . . . . . . . . . . . . 124

acute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

vertical . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

central . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

attractive fixed point . . . . . . . . . . . . . . . . . . 137

coterminal . . . . . . . . . . . . . . . . . . . . . . . . . 29

average speed . . . . . . . . . . . . . . . . . . . . . . . . . 100

general . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

included . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

C

initial side of . . . . . . . . . . . . . . . . . . . . . . . 24

Cartesian coordinates . . . . . . . . . . . . . . . . . . 25

inscribed . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

central angle . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

negative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

centripetal acceleration . . . . . . . . . . . . . . . 102

obtuse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

chord . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59, 94

of incidence . . . . . . . . . . . . . . . . . . . . . . . . 75

circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3, 6

of transmission . . . . . . . . . . . . . . . . . . . . 75

area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

positive . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

circumference . . . . . . . . . . . . . . . . . . . . . . 87

reference . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

circumscribed . . . . . . . . . . . . . . . . . . . . . . 59

right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

inscribed . . . . . . . . . . . . . . . . . . . . . . . 23, 62

straight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

segment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

subtended . . . . . . . . . . . . . . . . . . . . . . . . . . 95

tangent line to . . . . . . . . . . . . . . . . . . . . . . 6

terminal side of . . . . . . . . . . . . . . . . . . . . 24

unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

angle of depression . . . . . . . . . . . . . . . . . . . . . 15

circular motion . . . . . . . . . . . . . . . . . . . . . . . . 100

angle of elevation . . . . . . . . . . . . . . . . . . . . . . . 14

circumference . . . . . . . . . . . . . . . . . . . . . . . . . . 87

angular speed . . . . . . . . . . . . . . . . . . . . . . . . . 100

circumscribed circle . . . . . . . . . . . . . . . . . . . . 59

arc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

cis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

169

170

Index

Cofunction theorem . . . . . . . . . . . . . . . . . . . . 11

F

cofunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Fresnel coefficients. . . . . . . . . .75, 77, 85, 86

complementary angles . . . . . . . . . . . . . . . . . . . 2

function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

complex number . . . . . . . . . . . . . . . . . . . . . . 139

nth roots of. . . . . . . . . . . . . . . . . . . . . . . .144

G

argument of . . . . . . . . . . . . . . . . . . . . . . 141

gears . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

conjugate of . . . . . . . . . . . . . . . . . . . . . . . 140

gradian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

modulus of . . . . . . . . . . . . . . . . . . . . . . . . 140

graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

trigonometric form. . . . . . . . . . . . . . . .141

Greek alphabet . . . . . . . . . . . . . . . . . . . . . . . . . . 2

complex plane . . . . . . . . . . . . . . . . . . . . . . . . . 141

H

congruent triangles . . . . . . . . . . . . . . . . . . . . 17

half-angle formulas . . . . . . . . . . . . . . . . . . . . 79

conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

Heron’s formula . . . . . . . . . . . . . . . . . . . . . . . . 55

conjugate angles . . . . . . . . . . . . . . . . . . . . . . . . . 2

numerical instability of . . . . . . . . . . . . 57

coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Hipparchus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Cartesian . . . . . . . . . . . . . . . . . . . . . . . . . . 25

horizontal asymptote. . . . . . . . . . . . . . . . . . 124

polar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

hypotenuse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

rectangular . . . . . . . . . . . . . . . . . . . . . . . 146

cosecant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

I

graph of . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

difference of two angles . . . . . . . . . . . . 72

graph of . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

double-angle . . . . . . . . . . . . . . . . . . . . . . . 78

cotangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

half-angle . . . . . . . . . . . . . . . . . . . . . . . . . . 79

graph of . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

product-to-sum. . . . . . . . . . . . . . . . . . . . .82

coterminal angles . . . . . . . . . . . . . . . . . . . . . . 29

sum of two angles . . . . . . . . . . . . . . . . . . 71

cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

sum-to-product. . . . . . . . . . . . . . . . . . . . .82

imaginary number . . . . . . . . . . . . . . . . . . . . 139

D

pure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

De Moivre’s Theorem . . . . . . . . . . . . . . . . . 143

imaginary part . . . . . . . . . . . . . . . . . . . . . . . . 139

die punch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

imaginary unit . . . . . . . . . . . . . . . . . . . . . . . . 139

distance from earth to sun . . . . . . . . . . . . . 16

included angle . . . . . . . . . . . . . . . . . . . . . . . . . . 46

dodecagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

index of refraction . . . . . . . . . . . . . . . . . . . . . . 75

domain . . . . . . . . . . . . . . . . . . . . . . . . . . . 109, 120

initial side of angle . . . . . . . . . . . . . . . . . . . . . 24

double-angle formulas . . . . . . . . . . . . . . . . . . 78

inscribed angle . . . . . . . . . . . . . . . . . . . . . . . . . 59

inscribed circle . . . . . . . . . . . . . . . . . . . . . 23, 62

E

inscribed polygon . . . . . . . . . . . . . . . . . . . . . . . 23

ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

inscribed triangle . . . . . . . . . . . . . . . . . . . . . . 59

ellipticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

instantaneous center of rotation . . . . . . . 18

equatorial parallax . . . . . . . . . . . . . . . . . . . . . 16

inverse cosecant . . . . . . . . . . . . . . . . . . . . . . . 126

equilateral triangle . . . . . . . . . . . . . . . . . . . . . 10

inverse cosine . . . . . . . . . . . . . . . . . . . . . . . . . 122

Euclid’s formula . . . . . . . . . . . . . . . . . . . . . . . . . 6

inverse cotangent . . . . . . . . . . . . . . . . . . . . . 126

even function . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

inverse secant . . . . . . . . . . . . . . . . . . . . . . . . . 126

explementary angles . . . . . . . . . . . . . . . . . . . . 2

inverse sine . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

Index

171

inverse tangent . . . . . . . . . . . . . . . . . . . . . . . 124

Ptolemy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

inverse trigonometric functions . . . 19, 120

Ptolemy’s Theorem . . . . . . . . . . . . . . . . . . . . . 74

isosceles triangle . . . . . . . . . . . . . . . . . . . . . . . . 2

pure imaginary number . . . . . . . . . . . . . . . 139

Pythagorean Theorem . . . . . . . . . . . . . . . . . . . 4

L

Pythagorean triple . . . . . . . . . . . . . . . . . . . . . . 6

Law of Cosines . . . . . . . . . . . . . . . . . . . . . 38, 44

Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Q

ambiguous case . . . . . . . . . . . . . . . . . . . . 41

quadrants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Law of Tangents . . . . . . . . . . . . . . . . . . . . 38, 51

legs of a right triangle . . . . . . . . . . . . . . . . . . . 4

R

length of a line segment . . . . . . . . . . . . . . . . . 9

radians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

linear speed . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

radius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

radius of earth . . . . . . . . . . . . . . . . . . . . . . . . . 15

M

radius of sun . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

median of a triangle . . . . . . . . . . . . . . . . . . . . 49

range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109, 120

minute. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

real part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

modulated wave . . . . . . . . . . . . . . . . . . . . . . . 119

rectangular coordinates . . . . . . . . . . . . . . . 146

modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

reference angle . . . . . . . . . . . . . . . . . . . . . . . . . 29

Mollweide’s equations . . . . . . . . . . . . . . . . . . 52

reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . 32, 34

N

regular polygon. . . . . . . . . . . . . . . . . . . . . . . . .23

revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

negative angle . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Rhind Papyrus . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

nth roots of a complex number . . . . . . . . 144

right angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

numerical instability . . . . . . . . . . . . . . . . . . . 57

right triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

numerical methods . . . . . . . . . . . . . . . . . . . . 133

rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

O

rpm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

oblique triangle . . . . . . . . . . . . . . . . . . . . . . . . 38

S

obtuse angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

obtuse triangle . . . . . . . . . . . . . . . . . . . . . . . . . 38

secant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

octagon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

graph of . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

odd function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

secant method. . . . . . . . . . . . . . . . . . . . . . . . . 133

one-to-one . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

second . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

P

segment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

period of a function . . . . . . . . . . . . . . . . . . . 109

similar triangles . . . . . . . . . . . . . . . . . . . . . . . . . 4

perpendicular . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

sine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

perpendicular bisector. . . . . . . . . . . . . . . . . . 61

graph of . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

phase shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

unit circle definition of. . . . . . . . . . . . 103

phase, out of or in . . . . . . . . . . . . . . . . . . . . . 118

sine-Gordon equation . . . . . . . . . . . . . . . . . . 79

polar axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

sinusoidal curves . . . . . . . . . . . . . . . . . . . . . . 108

polar coordinates . . . . . . . . . . . . . . . . . . . . . . 146

slider-crank mechanism . . . . . . . . . . . . . . . . 18

positive angle. . . . . . . . . . . . . . . . . . . . . . . . . . . 24

slope of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

product-to-sum formulas . . . . . . . . . . . . . . . 82

Snell’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

172

Index

solving a triangle . . . . . . . . . . . . . . . . . . . . . . . 19

Y

standard position . . . . . . . . . . . . . . . . . . . . . . . 25

y-axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

straight angle. . . . . . . . . . . . . . . . . . . . . . . . . . . .1

y-intercept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

subtended angle . . . . . . . . . . . . . . . . . . . . . . . . 95

subtraction formulas . . . . . . . . . . . . . . . . . . . 72

sum-to-product formulas . . . . . . . . . . . . . . . 82

supplementary angles . . . . . . . . . . . . . . . . . . . 2

T

tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

graph of . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

tangent line . . . . . . . . . . . . . . . . . . . . . . 6, 15, 17

terminal side of angle . . . . . . . . . . . . . . . . . . 24

Thales’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 3

triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

acute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

equilateral . . . . . . . . . . . . . . . . . . . . . . . . . 10

inscribed . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

isosceles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

median of . . . . . . . . . . . . . . . . . . . . . . . . . . 49

oblique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

obtuse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

right . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

trigonometric equations . . . . . . . . . . . . . . . 129

trigonometric form . . . . . . . . . . . . . . . . . . . . 141

trigonometric functions . . . . . . . . . . . . . . . . . . 7

graphs of. . . . . . . . . . . . . . . . . . . . . . . . . .103

U

unit circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

V

v-block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

vertical asymptote . . . . . . . . . . . . . . . . . . . . 105

W

worm thread . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

X

x-axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Document Outline

- Preface

- Contents

- 1 Right Triangle Trigonometry

- 1.1 Angles

- 1.2 Trigonometric Functions of an Acute Angle

- 1.3 Applications and Solving Right Triangles

- 1.4 Trigonometric Functions of Any Angle

- 1.5 Rotations and Reflections of Angles

- 2 General Triangles

- 2.1 The Law of Sines

- 2.2 The Law of Cosines

- 2.3 The Law of Tangents

- 2.4 The Area of a Triangle

- 2.5 Circumscribed and Inscribed Circles

- 3 Identities

- 3.1 Basic Trigonometric Identities

- 3.2 Sum and Difference Formulas

- 3.3 Double-Angle and Half-Angle Formulas

- 3.4 Other Identities

- 4 Radian Measure

- 4.1 Radians and Degrees

- 4.2 Arc Length

- 4.3 Area of a Sector

- 4.4 Circular Motion: Linear and Angular Speed

- 5 Graphing and Inverse Functions

- 5.1 Graphing the Trigonometric Functions

- 5.2 Properties of Graphs of Trigonometric Functions

- 5.3 Inverse Trigonometric Functions

- 6 Additional Topics

- 6.1 Solving Trigonometric Equations

- 6.2 Numerical Methods in Trigonometry

- 6.3 Complex Numbers

- 6.4 Polar Coordinates

- Appendix A: Answers and Hints to Selected Exercises

- Appendix B: Graphing with Gnuplot

- GNU Free Documentation License

- History

- Index

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Table of Contents

Preface

1

1.2

1.3

1.4

1.5

2

2.2

2.3

2.4

2.5

3

3.2

3.3

3.4

4

4.2

4.3

4.4

5

5.2

5.3

6

6.2

6.3

6.4

Appendix A:

Appendix B:

GNU Free Documentation License

History

Index

1.1.3(

1.1.4

11)

1.2.1

:3

1.7.

1.2.2

1.2.3.

1.2.4,

.4

1.15

t5

1.3.2

1.3.3

9

1.3.5

1.3.6,

1.4.3(

1.4.4(