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’1 ¤ cos θ ¤ 1 ’ 1 ¤ sin θ ¤ 1.
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

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