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CHAPTER 6:
The Trigonometric Functions
6.1
6.2
6.3
6.4
6.5
6.6
The Trigonometric Functions of Acute Angles
Applications of Right Triangles
Trigonometric Functions of Any Angle
Radians, Arc Length, and Angular Speed
Circular functions: Graphs and Properties
Graphs of Transformed Sine and Cosine
Functions
Copyright © 2009 Pearson Education, Inc.
6.3
Trigonometric Functions of Any Angle





Find angles that are coterminal with a given angle and
find the complement and the supplement of a given
angle.
Determine the six trigonometric function values for
any angle in standard position when the coordinates of
a point on the terminal side are given.
Find the function values for any angle whose terminal
side lies on an axis.
Find the function values for an angle whose terminal
side makes an angle of 30º, 45º, or 60º with the x-axis.
Use a calculator to find function values and angles.
Copyright © 2009 Pearson Education, Inc.
Angle
An angle is the union of two rays with a common
endpoint called the vertex. We can think of it as a
rotation. Locate a ray along the positive x-axis with its
endpoint at the origin. This ray is called the initial side
of the angle. Now rotate a copy of this ray. A rotation
counterclockwise is a positive rotation and rotation
clockwise is a negative rotation. The ray at the end of
the rotation is called the terminal side of the angle.
The angle formed is said to be in standard position.
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 4
Angle
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Slide 6.3 - 5
Angle
The measure of an angle or rotation may be given in
degrees. One complete positive revolution or rotation
has a measure of 360º. One half of a revolution has a
measure of 180º …
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 6
Angle
One fourth of a revolution has a measure of 90º, and so
on.
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Slide 6.3 - 7
Angle
Angle measure of 60º, 135º, 330º, and 420º have terminal
sides that lie in quadrants I, II, IV and I respectively.
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 8
Angle
The negative rotations –30º, –110º, and –225º represent
angles with terminal sides in quadrants IV, III, and II
respectively.
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Slide 6.3 - 9
Coterminal Angles
If two or more angles have the same terminal side, the
angles are said to be coterminal. To find angles
coterminal with given angles, we add or subtract
multiples of 360º.
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 10
Example
Find two positive angles and two negative
angles that are coterminal with (a) 51º (b) –7º.
Solution:
a) Add or subtract multiples of 360º. Many
answers are possible.
51º + 360º = 411º
Copyright © 2009 Pearson Education, Inc.
51º + 3(360º) = 1131º
Slide 6.3 - 11
Example
Solution continued
51º – 360º = –309º
51º – 2(360º) = –669º
b) We have the following:
–7º + 360º = 353º
–7º + 2(360º) = 713º
–7º – 360º = –367º
–7º – 10(360º) = –3607º
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 12
Classification of Angles
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Slide 6.3 - 13
Complementary Angles
Two acute angles are complementary if their sum is
90º. For example, angles that measure 10º and 80º are
complementary because 10º + 80º = 90º.
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 14
Supplementary Angles
Two positive angles are supplementary if their sum is
180º. For example, angles that measure 45º and 135º
are supplementary because 45º + 135º = 180º.
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 15
Example
Find the complement and supplement of 71.46º.
Solution:
90º 71.46º  18.54º
180  71.46º  108.54º
The complement of 71.46º is 18.54º and the
supplement of 71.46º is 108.54º.
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 16
Trigonometric Functions of Angles
Consider a right triangle with
one vertex at the origin of a
coordinate system and one
vertex on the positive x-axis.
The other vertex P, a point on
the circle whose center is at the
origin and whose radius r is the
length of the hypotenuse of the
triangle. This triangle is a reference triangle for angle
, which is in standard position. Note that y is the
length of the side opposite  and x is the length of the
side adjacent to .
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 17
Trigonometric Functions of Angles
The three trigonometric functions of  are defined as
follows:
opp y
adj x
opp y
sin 

cos 

tan 

hyp r
hyp r
adj x
Since x and y are the coordinates of the point P and the
length of the radius is the hypotenuse, we have:
y-coordinate
sin  
radius
x-coordinate
cos 
radius
Copyright © 2009 Pearson Education, Inc.
y-coordinate
tan  
x-coordinate
Slide 6.3 - 18
Trigonometric Functions of Angles
We will use these definitions for functions of angles of
any measure.
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 19
Trigonometric Functions of Any Angle 
Suppose that P(x, y) is any point other than the vertex
on the terminal side of any angle  in standard
position, and r is the radius, or distance from the origin
to P(x,y). Then the trigonometric functions are defined
as follows:
y-coordinate y
sin 

radius
r
radius
r
csc 

y-coordinate y
x-coordinate x
cos 

radius
r
y-coordinate y
tan  

x-coordinate x
radius
sec 

x-coordinate
x-coordinate
cot  

y-coordinate
Copyright © 2009 Pearson Education, Inc.
r
x
x
y
Slide 6.3 - 20
Example
Find the six trigonometric
function values for the angle
shown.
Solution:
Determine r, distance from (0, 0) to (–4, –3).
r
2
2
x

0

y

0

x

y
   
r
4   3
2
2
Copyright © 2009 Pearson Education, Inc.
2
2
 16  9  25  5
Slide 6.3 - 21
Example
Solution continued
Substitute –4 for x, –3 for y, and 5 for r.
y 3
3
sin   

r
5
5
r 5
5
csc  

y 3
3
x 4
4
cos  

r
5
5
r
5
5
sec  

x 4
4
y 3 3
tan   

x 4 4
x 4 4
cot   

y 3 3
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Slide 6.3 - 22
Example
2
Given that tan   and  is in the second quadrant,
3
find the other function values.
Solution:
y
2 2
Sketch a second-quadrant angle using tan     
x
3 3
hyp  2 2  32  13
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Slide 6.3 - 23
Example
Solution continued
Use the lengths of the three sides to find the
appropriate ratios.
2
2 13
sin  

13
13
13
csc 
2
3
3 13
cos  

13
13
13
sec  
3
2
tan  
3
3
cot   
2
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 24
Terminal Side on an Axis
An angle whose terminal side falls on one of the axes
is a quadrantal angle. One of the coordinates of any
point on that side is 0. The definitions of the
trigonometric functions still apply, but in some cases,
function values will not be defined because a
denominator will be 0.
Copyright © 2009 Pearson Education, Inc.
Slide 6.3 - 25
Example
Find the sine, cosine, and tangent values for 90º, 180º,
270º, and 360º.
Solution:
Sketch the angle in standard position, label a point on
the terminal side, choosing (0, 1).
1
sin 90º   1
1
0
cos 90º   0
1
1
tan 90º  Not defined
0
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Slide 6.3 - 26
Example
Solution continued
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0
sin180º   0
1
1
cos180º 
 1
1
0
tan180º 
0
1
1
sin 270º 
 1
1
0
cos 270º   0
1
1
tan 270º 
Not defined
0
Slide 6.3 - 27
Example
Solution continued
0
sin 360º   0
1
1
cos 360º   1
1
0
tan 360º   0
1
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Slide 6.3 - 28
Reference Angles: 30º (45º and 60º)
Consider the angle 150º, its terminal side makes a 30º
angle with the x-axis.
1
sin150º  sin 30º 
2
3
cos150º   cos 30º  
2
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tan150º   tan 30º
1
3


3
3
Slide 6.3 - 29
Example
Find the sine, cosine, and tangent values for each of the
following:
a) 225º
b) –780º
Solution:
Draw the figure,
terminal side 225º,
reference angle is
225º – 180º = 45º
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Slide 6.3 - 30
Example
Solution continued
2
sin 225º  
2
2
cos 225º  
2
tan 225º  1
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Slide 6.3 - 31
Example
Solution continued
Draw the figure,
terminal side –780º is
coterminal with
–780º + 2(360º) = –60º,
reference angle is 60º.
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Slide 6.3 - 32
Example
Solution continued
3
sin 780º   
2
1
cos 780º  
2
tan 780º    3
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Slide 6.3 - 33
Example
Given the function value and the quadrant restriction,
find .
a) sin  = 0.2812, 90º <  < 180º
b) cot  = –0.1611, 270º <  < 360º
Solution:
Sketch the angle in the second quadrant.
Use a calculator to find the
acute (reference) angle
whose sine is 0.2812. It’s
approximately 16.33º. Now
180º – 16.33º = 163.37º.
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Slide 6.3 - 34
Example
Solution continued
b) cot  = –0.1611, 270º <  < 360º
Sketch the angle in the fourth quadrant.
1
1
tan 

 6.2073
cot  0.1611
Use a calculator to find the
acute (reference) angle
whose tangent is –6.2073.
It’s approximately 80.85º.
Now 360º – 80.85 = 279.15º.
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Slide 6.3 - 35
Bearing: Second-Type
In aerial navigation, directions, or bearings, are given
in degrees clockwise from north. Thus east is 90º,
south is 180º, and west is 270º.
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Slide 6.3 - 36
Example
An airplane flies 218
mi from an airport in a
direction of 245º.
How far south of the
airport is the plane
then? How far west?
Solution:
Sketch a diagram.
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Slide 6.3 - 37
Example
Solution continued:
Find the measure of angle ABC:
B  270º 245º  25º
Find how far south the plane is, that is, the length b:
b
 sin 25º
218
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b  218sin 25º  92 mi
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Example
Solution continued
Find how far west the plane is, that is, the length a:
a
 cos 25º
218
a  218 cos25º  198 mi
The airplane is about 92 mi south and about 198 mi
west of the airport.
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Slide 6.3 - 39
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