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Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 1
The Trigonometric
Functions
Chapter 5
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
5.1
Trigonometric Functions of
Acute Angles

Determine the six trigonometric ratios for a given acute
angle of a right triangle.

Determine the trigonometric function values of 30°, 45°,
and 60°.

Using a calculator, find function values for any acute
angle, and given a function value of an acute angle,
find the angle.

Given the function values of an acute angle, find the
function values of its complement.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Trigonometric Ratios

The figure illustrates how a right triangle is labeled with
reference to a given acute angle, .

The lengths of the sides of the triangle are used to
define the six trigonometric ratios:
sine (sin)
cosecant (csc)
cosine (cos)
secant (sec)
tangent (tan)
cotangent (cot)
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Slide 8- 4
Sine and Cosine

The sine of  is the length of the side opposite 
divided by the length of the hypotenuse:
Hypotenuse
side opposite 
sin  
hypotenuse


Side
Opposite

Side Adjacent
to 
The cosine of  is the length of the side adjacent to 
divided by the length of the hypotenuse.
side adjacent to 
cos 
hypotenuse
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Slide 8- 5
Trigonometric Function Values of an
Acute Angle 
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Slide 8- 6
Example

Use the triangle shown to calculate the six trigonometric
function values of .
Solution:
opp 24
sin  

hyp 25
adj
7
cos 

hyp 25
opp 24
tan  

adj
7
hyp 25
csc 

opp 24
hyp 25
sec 

adj
7
adj
7
cot  

opp 24
25
24

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7
Slide 8- 7
Reciprocal Functions

Reciprocal Relationships
1
csc 
sin 
1
sec 
cos
1
cot  
tan 
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Slide 8- 8
Pythagorean Theorem

The Pythagorean theorem may be used to find a
missing side of a right triangle.
2 2  52  h 2
4  25  h
29  h 2
29  h

h
2
5

2
This procedure can be combined with the reciprocal
relationships to find the six trigonometric function
values.
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Slide 8- 9
Example

5
If the tan   , find the other five trigonometric function
values of . 2
Solution: Find the length of the hypotenuse.
5
5 29
sin  

29
29
2
2 29
cos 

29
29
5
tan  
2
29
csc 
5
29
sec 
2
2
cot  
5
29
5

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2
Slide 8- 10
Function Values of 30 and 60



When the ratio of the opposite side to the hypotenuse is
½,  must have a measure of 30.
1
sin  
2
2
1

  30
Using the Pythagorean theorem the missing side is 3.
The missing angle must have a measure of 60.
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Slide 8- 11
Function Values of 30 and 60
60
2
1
30
3
1
sin 30 
2
3
cos30 
2
1
3
tan 30 

3
3
3
sin 60 
2
1
cos 60 
2
tan 60  3
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Slide 8- 12
Function Values of 45

The legs of this triangle must be equal, since they are
opposite congruent angles.

The hypotenuse is found by:
12  12  h 2
11  h
45
2
h
1
2  h2
2h
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45
1
Slide 8- 13
Function Values of 45 continued
1
2
sin 45 

2
2
1
2
cos 45 

2
2
1
tan 45   1
1
45
2
1
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45
1
Slide 8- 14
Summary of Function Values
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Slide 8- 15
Example

As a hot-air balloon began to rise, the ground crew
drove 1.2 mi to an observation station. The initial
observation from the station estimated the angle
between the ground and the line of sight to the balloon
to be 30. Approximately how high was the balloon at
that point? (We are assuming that the wind velocity was
low and that the balloon rose vertically for the first few
minutes.)
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Slide 8- 16
Example continued
Solution: We begin with a drawing of the situation. We
know the measure of an acute angle and the length of
its adjacent side.
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Slide 8- 17
Example continued

Since we want to determine the length of the opposite
side, we can use the tangent ratio, or the cotangent
ratio.
opp
h
tan 30 

adj 1.2
1.2 tan 30  h
 3
1.2 
h
 3 
0.7  h

The balloon is approximately 0.7 mi, or 3696 ft, high.
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Slide 8- 18
Cofunctions and Complements

The trigonometric function values for pairs of angles
that are complements have a special relationship. They
are called cofunctions.
sin   cos(90   )
cos  sin(90   )
tan   cot(90   )
cot   tan(90   )
sec  csc(90   )
csc  sec(90   )
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Slide 8- 19
Example

Given that sin 40
0.6428, cos 40  0.7660,
and
tan 40  0.8391,
find the six trigonometric
function values of 50.
1
csc 40 
 1.5557
sin 40
1
sec 40 
 1.3055
cos 40
1
cot 40 
 1.1918
tan 40
sin 50  cos 40  0.7660
tan 50  cot 40  1.1918
sec50  csc 40  1.5557
cos50  sin 40  0.6428
cot 50  tan 40  0.8391
csc50  sec 40  1.3055
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Slide 8- 20
5.2
Applications of Right Triangles

Solve right triangles.

Solve applied problems involving right triangles and
trigonometric functions.
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Solving Right Triangles

To solve a right triangle means to find the lengths of all
sides and the measures of all angles. This can be done
using right triangle trigonometry.
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Slide 8- 22
Example

In ABC , find a, b, and B.
Solution:
a
sin 42 
16.5
a  16.5sin 42
a  11.0
b
cos 42 
16.5
b  16.5cos 42
b  12.3
B
16.5
A
a
42
b
C
B = 90  42 = 48
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Slide 8- 23
Definitions

Angle of elevation: angle
between the horizontal
and a line of sight above
the horizontal.

Angle of depression:
angle between the
horizontal and a line of
sight below the
horizontal.
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Slide 8- 24
Example

To determine the height of a tree, a forester walks 100
feet from the base of the tree. From this point, he
measures the angle of elevation to the top of the tree to
be 47. What is the height of the tree?
h
tan 47 
100
h  100 tan 47
h  107.2 ft
h
47
100 ft
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Slide 8- 25
Bearing

Bearing is a method of giving directions. It involves
acute angle measurements with reference to a northsouth line.
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Slide 8- 26
Example

An airplane leaves the airport flying at a bearing of
N32W for 200 miles and lands. How far west of its
starting point is the plane?
w
200
w  200sin 32
w  106
w
sin 32 

200
32
The airplane is approximately 106 miles west of its
starting point.
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Slide 8- 27
5.3
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 30°, 45°, or 60° with the x-axis.

Use a calculator to find function values and angles.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Angle in Standard Position

An angle formed by it’s initial side along the positive
x-axis, with it’s vertex at the origin, and it’s terminal side
placed at the end of the rotation.
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Slide 8- 29
Coterminal Angles


Two or more angles that have the same terminal side.
For example, angles of measure 60 and 420 are
coterminal because they share the same terminal side.
Example: Find two positive and two negative angles
that are coterminal with 30.

30  360  390
30  360  330
30  2(360 )  750
30  2(360 )  690
390, 750, 330, and 690 are coterminal with 30.
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Slide 8- 30
Trigonometric Functions of Any Angle 
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Slide 8- 31
Example


Find the six trigonometric function values for the angle
shown:
Solution: First, determine r.
(2,4)
r
r  2 4
2
2

4
2
r  20
r  45  4  5
r2 5
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Slide 8- 32
Example continued

The six trigonometric functions values are:
sin  
y
4
2 5


r 2 5
5
x
2
5


r 2 5
5
y 4
tan     2
x 2
cos 
csc 
r 2 5
5


y
4
2
r 2 5
sec  
 5
x
2
x 2 1
cot    
y 4 2
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Slide 8- 33
Another Example

1
Given that tan   , and  is in the first quadrant, find
2
the other function values.
Solution: Sketch and label the angle. Find any missing
sides.
(2,1)
r
r  2 1
2
2

2
r  4 1
r 5
1
sin  
1
5

5
5
2
2 5
cos 

5
5
1
tan  
2
5
 5
1
5
sec 
2
2
cot    2
1
csc 
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Slide 8- 34
Reference Angle

The reference angle for an angle is the acute angle
formed by the terminal side of the angle and the x-axis.

The reference angle can be used when trying to find the
trigonometric function values for angles that cover more
than one quadrant. (ex. 210)
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Slide 8- 35
Example


Find the sine, cosine, and tangent function values for
210.
Solution: Draw the angle.
210
1


3
30
2
Note that there is a 30 angle in the third quadrant.
Label the sides of the triangle with 3, 1, and 2 as
shown.
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Slide 8- 36
Example continued
1
sin 210  
2
3
cos 210  
2
1
tan 210 
2

210
1
3
30
2
Notice that both the sine and cosine are negative
because the angle measuring 210 is in the third
quadrant.
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Slide 8- 37
5.4
Radians, Arc Length, and
Angular Speed

Find points on the unit circle determined by real
numbers.

Convert between radian and degree measure; find
coterminal, complementary, and supplementary
angles.

Find the length of an arc of a circle; find the measure of
a central angle of a circle.

Convert between linear speed and angular speed.
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Radian Measure

An angle measures 1 radian when the angle intercepts
an arc on a circle equal to the radius of the circle.

1 radian is approximately 57.3.
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Slide 8- 39
Converting between Degree and Radian
Measure
 radians
180

To convert from degree to radian measure, multiply by
 radians
180

180

 1.
 radians
.
To convert from radian to degree measure, multiply by
180
.
 radians
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Slide 8- 40
Example

Convert each of the following to either radians or
degrees.
a) 150
b) 75

150 5
150 


radians
180 180
6
c) 7 radians
4
7 180 1260


 315
4 
4

75
5
75 


180 180
12
d) 3 radians
3
180


540

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 171.9
Slide 8- 41
Radian Measure
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Slide 8- 42
Example

Find the length of an arc of a circle of radius 10 cm

associated with an angle of
radians.
4
s

r
s  r
or
s  r

10 5
s  10  

4
4
2
s  7.85 cm
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Slide 8- 43
Definitions

Linear Speed: the distance traveled per unit of time,
where s is the distance and t is the time.
s
v
t

Angular Speed: the amount of rotation per unit of time,
where  is the angle of rotation and t is the time.


t
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Slide 8- 44
Linear Speed in Terms of Angular Speed

The linear speed v of a point a distance r from the
center of rotation is given by v = r, where  is the
angular speed in radians per unit of time.
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Slide 8- 45
Example


Find the angle of revolution of a point on a circle of
diameter 30 in. if the point moves 4 in. per second for
11 seconds.
Since t = 11,  must be determined before we can solve
for .
v
v  r ,   .
r
4in./ sec

 0.26 per second
15in.
4
44
   t  11 
The angle of revolution of
15
15
the point is approximately
  2.93
3 radians.
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Slide 8- 46
5.5
Circular Functions:
Graphs and Properties

Given the coordinates of a point on the unit circle, find
its reflections across the x-axis, the y-axis, and the
origin.

Determine the six trigonometric function values for a
real number when the coordinates of the point on the
unit circle determined by that real number are given.

Find function values for any real number using a
calculator.

Graph the six circular functions and state their
properties.
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Basic Circular Functions
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Slide 8- 48
Example

Find each of the following function values.
2
3
3

c) tan
4
b) cos    
a) sin
 4
5

d) sec
6
Solutions:
 1 3 
 ,

 2 2 
2
3
a) The coordinates of the point
2
determined by
are  1 , 3 
3
 2 2 
2
3
sin
y
3
2
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Slide 8- 49
Example continued

b) The coordinates of the point determined by 
4
are  2  2 

2
,


2 
 2


cos     x 
2
 4
3
c) The coordinates of the point determined by
4
are   2 2 
,

 tan 3  y  2 2  1
2 
 2
4
x  2 2
5
d) The coordinates of the point determined by
6
are   3 1 
, 

 2 2
5 1
1
2 3
sec
 

6
x  3 2
3
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Slide 8- 50
Graphs of the Sine and Cosine Functions
1. Make a table of values.
2. Plot the points.
3. Connect the points with a smooth curve.
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Slide 8- 51
Example
1. Make a table of values.
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Slide 8- 52
Example continued
2. Plot the points.
3. Connect the
points with
a smooth curve.
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Slide 8- 53
Domain and Range of Sine and Cosine
Functions

The domain of the sine and cosine functions is
(, ).
The range of the sine and cosine functions is [1, 1].

Periodic Function

A function f is said to be periodic if there exists a
positive constant p such that f(s + p) = f(s) for all s in the
domain of f. The smallest such positive number p is
called the period of the function.
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Slide 8- 54
Amplitude

The amplitude of a periodic function is defined as one
half of the distance between its maximum and minimum
function values.

The amplitude is always positive.

The amplitude of y = sin x and y = cos x is 1.
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Slide 8- 55
Graph of y = tan s
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Slide 8- 56
Graph of y = cot s
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Slide 8- 57
Graph of y = csc s
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Slide 8- 58
Graph of y = sec s
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Slide 8- 59
5.6
Graphs of Transformed Sine
and Cosine Functions

Graph transformations of y = sin x and y = cos x in the
form y = A sin (Bx – C) + D and y = A cos (Bx – C) + D
and determine the amplitude, the period, and the phase
shift.

Graph sums of functions.
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Graphs of Transformed Sine and Cosine
Functions: Vertical Translation

y = sin x + D and y = cos x + D

The constant D translates the graphs D units up if
D > 0 or |D| units down if D < 0.
Example:
Sketch a graph of
y = sin x  2.
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Slide 8- 61
Graphs of Transformed Sine and Cosine
Functions: Amplitude

y = A sin x and y = A cos x

If |A| > 1, then there will be a vertical stretching by a
factor of |A|.

If |A| < 1, then there will be a vertical shrinking by a
factor of |A|.

If A < 0, the graph is also reflected across the x-axis.
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Slide 8- 62
Example

Sketch a graph of
y = 3 sin x.

The sine graph
(y = sin x) is stretched
vertically by a factor of 3.
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Slide 8- 63
Graphs of Transformed Sine and Cosine
Functions: Period

y = sin Bx and y = cos Bx

If |B| < 1, then there will be a horizontal stretching.

If |B| > 1, then there will be a horizontal shrinking.

If B < 0, the graph is also reflected across the y-axis.

2
.
The period will be
B
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Slide 8- 64
Example

Sketch a graph of
y = sin 2x.

The sine graph
(y = sin x) is shrunk
horizontally.

2
The period is

2
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Slide 8- 65
Graphs of Transformed Sine and Cosine
Functions: Horizontal Translation or
Phase Shift

y = sin (x  C) and y = cos (x  C)

The constant C translates the graph horizontally
|C| units to the right if C > 0 and |C| units to the left if
C < 0.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 66
Example

Sketch the graph of
y = sin (x + ).

The sine graph
(y = sin x) is translated
 units to the left.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 67
Combined Transformations

y = A sin (Bx  C) + D and y = A cos (Bx  C) + D

The amplitude is |A|.

The period is

The graph is translated vertically D units.

The graph is translated horizontally C units.
2
.
B
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 68
Example


Find the vertical shift,
amplitude, period, and
phase shift for the
following function:
y = 2 sin (4x  2)  3.
Solution: Write the
function in standard
form.
y  2sin(4 x  2 )  3

|A| = |2| = 2 means the
amplitude is 2

B = 4 means the period
2 

is
 
2
y  2sin  4  x 
4
 


   (3)

4
2

 C/B = 2 means the phase
shift is  units to the right.
2
D = 3 means the vertical
shift is 3 units down.
 
 
y  2sin  4  x     (3)
2 
 
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 69
Example continued

 
 
Sketch y  2sin  4  x     (3)
2 
 
Amplitude = 2
Vertical shift = 3 down

Phase shift = 2 right

Period =

First, sketch y = sin 4x.




2
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 70
Example continued

Second, sketch
y = 2 sin 4x.

Third, sketch
 
 
y  2sin  4  x   
2 
 
.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 71
Example continued

Finally, sketch
 
 
y  2sin  4  x     (3) .
2 
 
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 8- 72