Transcript StewartPCalc6_06_03

Trigonometric Functions:
Right Triangle Approach
Copyright © Cengage Learning. All rights reserved.
Trigonometric Functions
6.3
of Angles
Copyright © Cengage Learning. All rights reserved.
Objectives
► Trigonometric Functions of Angles
► Evaluating Trigonometric Functions at Any
Angle
► Trigonometric Identities
► Areas of Triangles
3
Trigonometric Functions
of Angles
4
Reference Angle
Figure 6 shows that to find a reference angle it’s useful
to know the quadrant in which the terminal side of the angle
 lies.
The reference angle
for an angle 
Figure 5
5
Example 1 – Finding Reference Angles
Find the reference angle for
(a)
and
(b)  = 870°.
Solution:
(a) The reference angle is the acute angle formed by the
terminal side of the angle 5 /3 and the x-axis
(see Figure 7).
Figure 7
6
Example 1 – Solution
Since the terminal side of this angle is in Quadrant IV,
the reference angle is
(b) The angles 870° and 150° are coterminal
[because 870 – 2(360) = 150].
7
Example 1 – Solution
cont’d
Thus, the terminal side of this angle is in Quadrant II (see
Figure 8).
Figure 8
So the reference angle is
= 180° – 150° = 30°
8
Definitions of the Trigonometric Functions in Terms of
a Unit Circle
9
Example 2 – Finding Trigonometric Functions of Angles
Find (a) cos 135 and (b) tan 390.
Solution:
(a) From Figure 4 we see that cos 135 = x.
we have,
cos 135 =
Figure 4
10
Example 2 – Solution
cont’d
(b) The angles 390° and 30° are coterminal. From
Figure 5
it’s clear that tan 390° = tan 30° and, since tan 30°
=
Figure 5
we have,
tan 390° =
11
Example 3 – Using the Reference Angle to Evaluate
Trigonometric Functions
Find
(a)
and
(b)
Solution:
(a) The angle 16 /3 is coterminal with 4 /3, and these
angles are in Quadrant III (see Figure 11).
is negative
Figure 11
12
Example 3 – Solution
Thus, the reference angle is (4 /3) –  =  /3.
Since the value of sine is negative in Quadrant III, we have
13
Example 3 – Solution
cont’d
(b) The angle – /4 is in Quadrant IV, and its reference
angle is  /4 (see Figure 12).
is positive
is positive
Figure 12
14
Example 3 – Solution
cont’d
Since secant is positive in this quadrant, we get
15
Evaluating Trigonometric Functions
at Any Angles
16
Trigonometric Functions of Angles
Let POQ be a right triangle with acute angle  as shown in
Figure 1(a). Place  in standard position as shown in Figure
1(b).
(a)
(b)
Figure 1
17
Trigonometric Functions of Angles
Then P = P (x, y) is a point on the terminal side of . In
triangle POQ, the opposite side has length y and the
adjacent side has length x.
Using the Pythagorean Theorem, we see that the
hypotenuse has length r =
So
The other trigonometric ratios can be found in the same
way. These observations allow us to extend the
trigonometric ratios to any angle.
18
Trigonometric Functions of Angles
We define the trigonometric functions of angles as follows
(see Figure 2).
Figure 2
19
Example: Evaluating Trigonometric Functions
Let P = (1, –3) be a point on the terminal side of  Find each
of the six trigonometric functions of 
P = (1, –3) is a point on the terminal side of 
x = 1 and y = –3
r  x 2  y 2  (1)2  (3)2  1  9  10
y
3
-3 10
3 10
sin   
=
i
=r
10
10
10 10
1
1
10
10
x
=
i
=
cos  
r
10
10 10 10
20
Example: Evaluating Trigonometric Functions
(continued)
Let P = (1, –3) be a point on the terminal side of
of the six trigonometric functions of 
We have found that
 Find each
r  10.
y 3
 3
tan   
x 1
x
r
10
1
cot    
csc   
y
y
3
3
10
r
 10
sec  
1
x
21
Example: Evaluating Trigonometric Functions
(continued)
Let P = (1, –3) be a point on the terminal side of
of the six trigonometric functions of 
 Find each
3 10
sin   
10
10
csc  
3
10
cos 
10
sec  10
tan   3
1
cot   
3
22
Trigonometric Functions of Angles
The angles for which the trigonometric functions may be
undefined are the angles for which either the x- or
y-coordinate of a point on the terminal side of the angle
is 0.
These are quadrantal angles—angles that are coterminal
with the coordinate axes.
23
Example: Trigonometric Functions of Quadrantal
Angles
Evaluate, if possible, the cosine function and the cosecant
function at the following quadrantal angle:
  0  0
If   0  0 radians,then the terminal side of the angle
is on the positive x-axis. Let us select the point
P = (1, 0) with x = 1 and y = 0.
x 1
cos    1
r 1
r
1
csc  
csc is undefined.
y 0
24
Example: Trigonometric Functions of Quadrantal
Angles
Evaluate, if possible, the cosine function and the cosecant

function at the following quadrantal angle:   90 

2
radians, then the terminal side of the angle
If   90 
2
is on the positive y-axis. Let us select the point
P = (0, 1) with x = 0 and y = 1.
x 0
cos    0
r 1
r 1
csc    1
y 1
25
Example: Trigonometric Functions of Quadrantal
Angles
Evaluate, if possible, the cosine function and the cosecant
function at the following quadrantal angle:   180  
If   180   radians,
then the terminal side of the angle is
on the positive x-axis. Let us select the point
P = (–1, 0) with x = –1 and y = 0.
x 1
cos    1
r 1
r
1
csc  
y 0
csc is undefined.
26
Example: Trigonometric Functions of Quadrantal
Angles
Evaluate, if possible, the cosine function and the cosecant
function at the following quadrantal angle:   270  3
3
2
If   270 
radians,then the terminal side of the
2
angle is on the negative
y-axis. Let us select the point
P = (0, –1) with x = 0 and y = –1.
x 0
cos    0
r 1
r 1
csc    1
y 1
27
Evaluating Trigonometric Functions at Any Angle
From the definition we see that the values of the
trigonometric functions are all positive if the angle  has its
terminal side in Quadrant I.
This is because x and y are positive in this quadrant. [Of
course, r is always positive, since it is simply the distance
from the origin to the point P (x, y).] If the terminal side of 
is in Quadrant II, however, then x is negative and y is
positive.
Thus, in Quadrant II the functions sin  and csc  are
positive, and all the other trigonometric functions have
negative values.
28
Evaluating Trigonometric Functions at Any Angle
The following mnemonic device can be used to remember
which trigonometric functions are positive in each quadrant:
All of them, Sine, Tangent, or Cosine.
You can remember this as “All Students Take Calculus” or
“A Smart Trig Class.”
29
Evaluating Trigonometric Functions at Any Angle
You can check the other entries in the following table.
30
Evaluating Trigonometric Functions at Any Angle
31
Example: Evaluating Trigonometric Functions
1
Given tan    and cos  0, find sin q and secq .
3
Because both the tangent and the cosine are negative, 
lies in Quadrant II.
y
1
x  3, y  1
tan   
x 3
r  x 2  y 2  (3)2  (1)2  9  1  10
y
1
10
10
sin   =
i
=
r
10 10 10
10
10
r

sec  
3
x 3
32
Use x, y, and r.
33
Trigonometric Identities
34
Trigonometric Identities
The trigonometric functions of angles are related to each
other through several important equations called
trigonometric identities.
We’ve already encountered the reciprocal identities.
These identities continue to hold for any angle , provided
that both sides of the equation are defined.
35
Fundamental Identities
36
Example: Using Quotient and Reciprocal Identities
Given sin t 
2 and
5 find the value of each of the four
cos t 
3
3
remaining trigonometric functions.
2
2 5 2 5
2 3
2
sin t  3
= i =
= i =
tan t 
5
5
3 5
cos t
5 5
5
3
1 3
1
 
csc t 
sin t 2 2
3
37
Example: Using Quotient and Reciprocal Identities
(continued)
2
5 find the value of each of the four
Given sin t  and cos t 
3
3
remaining trigonometric functions.
3 5 3 5
1
3
1
= i =


sec t 
5
cos t
5
5
5 5
3
1
5
1
5
5 5 5
5


cot t 
=
i =
=
tan t
2 5 2 5 2 5 5 2i5 2
5
38
The Pythagorean Identities
39
Example: Using a Pythagorean Identity
Given that sin t 
1 and
 find the value of
0t  ,
2
2
cost using a trigonometric identity.
sin t  cos t  1
2
2
2
1
cos t  1 
4
2
 1   cos 2 t  1
 
2
3
cos t 
4
1
 cos 2 t  1
4
3
3
cos t 

4 2
2
40
Use trigonometric identities
41
Trigonometric Identities
The Pythagorean identities are a consequence of the
Pythagorean Theorem.
42
Example 5 – Expressing One Trigonometric Function in Terms of Another
(a) Express sin  in terms of cos .
(b) Express tan  in terms of sin , where  is in Quadrant II.
Solution:
(a) From the first Pythagorean identity we get
sin  =
where the sign depends on the quadrant.
43
Example 5 – Solution
cont’d
If  is in Quadrant I or II, then sin  is positive, so
sin  =
whereas if  is in Quadrant III or IV, sin  is negative, so
sin  =
44
Example 5 – Solution
cont’d
(b) Since tan  = sin  /cos , we need to write cos  in
terms of sin . By part (a)
cos  =
and since cos  is negative in Quadrant II, the negative
sign applies here. Thus
tan  =
45
Use trigonometric identities
46
The Domain and Range of the Sine and Cosine
Functions
y = cos x
D: ( -¥, ¥)
R: [-1, 1]
y = sin x
D: ( -¥, ¥)
R: [-1, 1]
47
The Domain and Range of Tangent Function
y = tan x
p
D: all reals except odd multiples of
2
R: ( -¥, ¥)
48
Even and Odd Trigonometric Functions
Even functions show y-axis symmetry.
Odd functions show origin symmetry.
49
Example: Using Even and Odd Functions to
Find Values of Trigonometric Functions
Find the value of each trigonometric function:




sec     sec    2
 4
4


2


sin      sin    
2
 4
4
50
Cofunction Identities
51
Using Cofunction Identities
Find a cofunction with the same value as the given expression:
a.
sin 46  cos(90  46)  cos 44
b.
 
6  
5


   tan
cot  tan     tan 
12
 2 12 
 12 12 
12

52
Trigonometric Identities
The Pythagorean identities are a consequence of the
Pythagorean Theorem.
53
Example 5 – Expressing One Trigonometric Function in Terms of Another
(a) Express sin  in terms of cos .
(b) Express tan  in terms of sin , where  is in Quadrant II.
Solution:
(a) From the first Pythagorean identity we get
sin  =
where the sign depends on the quadrant.
54
Example 5 – Solution
cont’d
If  is in Quadrant I or II, then sin  is positive, and so
sin  =
whereas if  is in Quadrant III or IV, sin  is negative,
and so
sin  =
55
Example 5 – Solution
cont’d
(b) Since tan  = sin  /cos , we need to write cos  in
terms of sin . By part (a)
cos  =
and since cos  is negative in Quadrant II, the negative
sign applies here. Thus
tan  =
56
Areas of Triangles
57
Areas of Triangles
The area of a triangle is
 base  height .
If we know two sides and the included angle of a triangle,
then we can find the height using the trigonometric
functions, and from this we can find the area.
If  is an acute angle, then the height of the triangle in
Figure 16(a) is given by h = b sin  . Thus the area is
 base  height
=
sin 
(a)
Figure 16
58
Areas of Triangles
If the angle  is not acute, then from Figure 16(b) we see
that the height of the triangle is
h = b sin(180° –  ) = b sin 
(b)
Figure 16
59
Areas of Triangles
This is so because the reference angle of  is the angle
180 – . Thus, in this case also, the area of the triangle is
 base  height =
sin 
60
Example 8 – Finding the Area of a Triangle
Find the area of triangle ABC shown in Figure 17.
Figure 17
Solution:
The triangle has sides of length 10 cm and 3 cm, with
included angle 120.
61
Example 8 – Solution
Therefore
sin 
(3) sin 120
= 15 sin 60
Reference angle
 13 cm2
62
Trigonometric Functions of Angles
It is a crucial fact that the values of the trigonometric
functions do not depend on the choice of the point P(x, y).
This is because if P(x , y ) is any other point on the
terminal side, as in Figure 3, then triangles POQ and
POQ are similar.
Figure 3
63
Evaluating Trigonometric Functions at Any Angle
64
Evaluating Trigonometric Functions at Any Angle
Figure 6 shows that to find a reference angle it’s useful
to know the quadrant in which the terminal side of the angle
 lies.
The reference angle
for an angle 
Figure 5
65