and

Math Note: It is common to write sin2 θ to mean (sin θ )2 and cos2 θ to mean

(cos θ )2 .

EXAMPLE 1.18

Compute the sine and cosine of π/3.

24 CHAPTER 1 Basics

SOLUTION

We sketch the terminal radius and associated triangle (see Fig. 1.33). This is

√

a 30“60“90 triangle whose sides have ratios 1 : 3 : 2. Thus

1 1

=2 x= .

or

2

x

Likewise,

√

√ √ 3

y

=3 y = 3x =

or .

2

x

It follows that

√

3

π

sin =

3 2

and

1

π

=.

cos

3 2

y

unit circle

√3

2

F/3

x

1

2

Fig. 1.33

You Try It: The cosine of a certain angle is 2/3. The angle lies in the fourth

quadrant. What is the sine of the angle?

Math Note: Notice that if θ is an angle then θ and θ + 2π have the same terminal

radius and the same terminal point (for adding 2π just adds one more trip around

the circle”look at Fig. 1.34).

As a result,

sin θ = x = sin(θ + 2π )

CHAPTER 1 Basics 25

y

sin G

G

cos G x

unit circle

Fig. 1.34

and

cos θ = y = cos(θ + 2π ).

We say that the sine and cosine functions have period 2π : the functions repeat

themselves every 2π units.

In practice, when we calculate the trigonometric functions of an angle θ, we

reduce it by multiples of 2π so that we can consider an equivalent angle θ , called

the associated principal angle, satisfying 0 ¤ θ < 2π . For instance,

15π/2 has associated principal angle

(since 15π/2 ’ 3π/2 = 3 · 2π )

3π/2

and

’10π/3 has associated principal angle

(since ’ 10π/3 ’ 2π/3 = ’12π/3 = ’2 · 2π ).

2π/3

You Try It: What are the principal angles associated with 7π , 11π/2, 8π/3,

’14π/5, ’16π/7?

What does the concept of angle and sine and cosine that we have presented here

have to do with the classical notion using triangles? Notice that any angle θ such

that 0 ¤ θ < π/2 has associated to it a right triangle in the ¬rst quadrant, with

vertex on the unit circle, such that the base is the segment connecting (0, 0) to (x, 0)

and the height is the segment connecting (x, 0) to (x, y). See Fig. 1.35.

26 CHAPTER 1 Basics

y

unit circle (x, y)

opposite

side

G

x

G

adjacent side

Fig. 1.35

Then

opposite side of triangle

y

sin θ = y = =

1 hypotenuse

and

adjacent side of triangle

x

cos θ = x = = .

1 hypotenuse

Thus, for angles θ between 0 and π/2, the new de¬nition of sine and cosine using

the unit circle is clearly equivalent to the classical de¬nition using adjacent and

opposite sides and the hypotenuse. For other angles θ, the classical approach is to

reduce to this special case by subtracting multiples of π/2. Our approach using the

unit circle is considerably clearer because it makes the signatures of sine and cosine

obvious.

Besides sine and cosine, there are four other trigonometric functions:

sin θ

y

tan θ = =

cos θ

x

cos θ

x

cot θ = =

sin θ

y

1 1

sec θ = =

cos θ

x

1 1

csc θ = = .

sin θ

y

Whereas sine and cosine have domain the entire real line, we notice that tan θ and

sec θ are unde¬ned at odd multiples of π/2 (because cosine will vanish there) and

cot θ and csc θ are unde¬ned at even multiples of π/2 (because sine will vanish

there). The graphs of the six trigonometric functions are shown in Fig. 1.36.

EXAMPLE 1.19

Compute all the trigonometric functions for the angle θ = 11π/4.

CHAPTER 1 Basics 27

1 1

0.5 0.5

_6 _4 _2 _6 _4 _2

2 4 6 2 4 6

_ 0.5 _ 0.5

_1 _1

Graphs of y = sin x and y = cos x.

Fig. 1.36(a)

30 30

20 20

10 10

_6 _4 _2 _6 _4 _2

2 4 6 2 4 6

_ 10 _ 10

_ 20 _ 20

_ 30 _ 30

Graphs of y = tan x and y = cot x.

Fig. 1.36(b)

15 15

10 10

5 5

_4 _2 _2 6

2 4 4

_4

_6 _6

6 2

_5 _5

_10 _10

_15 _15

Graphs of y = sec x and y = csc x.

Fig. 1.36(c)

SOLUTION

We ¬rst notice that the principal associated angle is 3π/4, so we deal with

that angle. Figure 1.37 shows that the triangle associated to this angle is an

isosceles right triangle with hypotenuse 1.

28 CHAPTER 1 Basics

y

11F/4

1

√2

x

1

√2

unit circle

Y Fig. 1.37

FL

AM

√ √

Therefore x = ’1/ 2 and y = 1/ 2. It follows that

1

sin θ = y = √

TE

2

1

cos θ = x = ’ √

2

y

tan θ = = ’1

x

x

cot θ = = ’1

y

√

1

sec θ = =’ 2

x

1√

csc θ = = 2.

y

Similar calculations allow us to complete the following table for the values of

the trigonometric functions at the principal angles which are multiples of π/6

or π/4.

CHAPTER 1 Basics 29

Angle Sin Cos Tan Cot Sec Csc

0 0 1 0 undef 1 undef

√ √ √ √

1/2 3/2 1/ 3 3 2/ 3 2

π/6

√ √ √ √

2/2 2/2 1 1 2 2

π/4

√ √ √ √

3/2 1/2 3 1/ 3 2 2/ 3

π/3

1 0 undef 0 undef 1

π/2

√ √ √ √

’1/2 ’3 ’1/ 3 ’2

2π/3 3/2 2/ 3

√ √ √ √

’ 2/2 ’1 ’1 ’2

3π/4 2/2 2

√ √ √ √

’ 3/2 ’1/ 3 ’3 ’2/ 3

5π/6 1/2 2

’1 ’1

0 0 undef undef

π

√ √ √ √

’1/2 ’ 3/2 ’2/ 3 ’2

7π/6 1/ 3 3

√ √ √ √

’ 2/2 ’ 2/2 ’2 ’2

5π/4 1 1

√ √ √ √

’ 3/2 ’1/2 ’2 ’2/ 3

4π/3 3 1/ 3

’1 ’1

3π/2 0 undef 0 undef

√ √ √ √

’ 3/2 ’3 ’1/ 3 ’2/ 3

5π/3 1/2 2

√ √ √ √

’ 2/2 ’1 ’1 ’2

7π/4 2/2 2

√ √ √ √

’1/2 ’1/ 3 ’3 ’2

11π/6 3/2 2/ 3

Besides properties (1) and (2) above, there are certain identities which are

fundamental to our study of the trigonometric functions. Here are the principal ones:

tan2 θ + 1 = sec2 θ

(3)

cot 2 θ + 1 = csc2 θ

(4)

sin(θ + ψ) = sin θ cos ψ + cos θ sin ψ

(5)

cos(θ + ψ) = cos θ cos ψ ’ sin θ sin ψ

(6)

sin(2θ ) = 2 sin θ cos θ

(7)

cos(2θ ) = cos2 θ ’ sin2 θ

(8)

sin(’θ ) = ’ sin θ

(9)

cos(’θ ) = cos θ

(10)

1 ’ cos 2θ

(11) sin2 θ =

2

1 + cos 2θ

(12) cos2 θ =

2

EXAMPLE 1.20

Prove identity number (3).

30 CHAPTER 1 Basics

SOLUTION

We have

sin2 θ

tan θ + 1 = +1

2

cos2 θ

sin2 θ cos2 θ

= +

cos2 θ cos2 θ

sin2 θ + cos2 θ

=

cos2 θ

1

=

cos2 θ

(where we have used Property (1))

= sec2 θ.

You Try It: Use identities (11) and (12) to calculate cos(π/12) and sin(π/12).

1.8 Sets and Functions

We have seen sets and functions throughout this review chapter, but it is well to

bring out some of the ideas explicitly.

A set is a collection of objects. We denote a set with a capital roman letter, such

as S or T or U . If S is a set and s is an object in that set then we write s ∈ S and

we say that s is an element of S. If S and T are sets then the collection of elements

common to the two sets is called the intersection of S and T and is written S © T .

The set of elements that are in S or in T or in both is called the union of S and T

and is written S ∪ T .

A function from a set S to a set T is a rule that assigns to each element of S a

unique element of T . We write f : S ’ T .

EXAMPLE 1.21

Let S be the set of all people who are alive at noon on October 10, 2004

and T the set of all real numbers. Let f be the rule that assigns to each

person his or her weight in pounds at precisely noon on October 10, 2004.

Discuss whether f : S ’ T is a function.

SOLUTION

Indeed f is a function since it assigns to each element of S a unique element

of T . Notice that each person has just one weight at noon on October 10, 2004:

that is a part of the de¬nition of “function.” However two different people may

have the same weight”that is allowed.

CHAPTER 1 Basics 31

EXAMPLE 1.22

Let S be the set of all people and T be the set of all people. Let f be the