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Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 1
Trigonometric Identities,
Inverse Functions,
and Equations
Chapter 6
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
6.1
Identities: Pythagorean and
Sum and Difference

State the Pythagorean identities.

Simplify and manipulate expressions containing
trigonometric expressions.

Use the sum and difference identities to find function
values.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Basic Identities

An identity is an equation that is true for all possible
replacements of the variables.
1
sin x 
,
csc x
1
cos x 
,
sec x
1
tan x 
,
cot x
1
csc x 
,
sin x
1
sec x 
,
cos x
1
cot x 
,
tan x
sin( x)   sin x,
cos( x)  cos x,
tan( x)   tan x,
sin x
tan x 
,
cos x
cos x
cot x 
sin x
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 4
Pythagorean Identities
sin x  cos x  1,
2
2
1  cot x  csc x,
2
2
1  tan 2 x  sec 2 x
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Slide 6- 5
Example

Multiply and simplify:
a) sin x(cot x  csc x)
Solution:
sin x(cot x  csc x)
 sin x cot x  sin x csc x
cos x
1
 sin x
 sin x
sin x
sin x
 cos x  1
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Slide 6- 6
Example continued
b) Factor and simplify:
Solution:
sin 4 x  sin 2 x cos 2 x
sin x  sin x cos x
4
2
2
 sin 2 x(sin 2 x  cos 2 x)
 sin 2 x  (1)
 sin 2 x
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Slide 6- 7
Another Example

Simplify the following
trigonometric expression:
Solution:
cos x
cos x

1  sin x 1  sin x
cos x(1  sin x)
cos x(1  sin x)

(1  sin x)(1  sin x) (1  sin x)(1  sin x)
cos x  sin x cos x cos x  sin x cos x


2
1  sin 2 x
1  sin x
2cos x

1  sin 2 x
2cos x

cos 2 x
2
or 2sec x

cos x
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Slide 6- 8
Sum and Difference Identities

There are six identities here, half of them obtained by
using the signs shown in color.
sin(u  v)  sin u cos v  cos u sin v,
cos(u  v)  cos u cos v sin u sin v,
tan u  tan v
tan(u  v) 
1 tan u tan v
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Slide 6- 9
Example

Find sin 75 exactly.
sin 75  sin(30  45 )
 sin 30 cos 45  cos30 sin 45
1 2
 
2 2
2


4
2

4
3 2


2 2
6
4
6
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Slide 6- 10
6.2
Identities: Cofunction,
Double-Angle, and Half-Angle

Use cofunction identities to derive other identities.

Use the double-angle identities to find function values of
twice an angle when one function value is known for
that angle.

Use the half-angle identities to find function values of
half an angle when one function value is known for that
angle.

Simplify trigonometric expressions using the doubleangle and half-angle identities.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Cofunction Identities




sin   x   cos x,
2



tan   x   cot x,
2



cos   x   sin x,
2



cot   x   tan x,
2



sec   x   csc x,
2



csc   x   sec x
2

Cofunction Identities for the Sine and Cosine


sin  x     cos x
2



cos  x    sin x
2

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Slide 6- 12
Example



Find an identity for cot  x   .
2

Solution:


cos  x  

2


cot  x   

2


sin  x  
2

sin x

 cos x
  tan x
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Slide 6- 13
Double-Angle Identities
sin 2 x  2sin x cos x,
2 tan x
tan 2 x 
1  tan 2 x
cos 2 x  cos 2 x  sin 2 x
 1  2sin 2 x
 2cos 2 x  1
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Slide 6- 14
Example

Find an equivalent expression for cos 3x.
Solution:
cos3 x  cos(2 x  x)
 cos 2 x cos x  sin 2 x sin x
 (1  2sin 2 x)cos x  2sin x cos x sin x
 cos x  2sin 2 x cos x  2sin 2 x cos x
 cos x  4sin 2 x cos x
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Slide 6- 15
Half-Angle Identities
x
1  cos x
sin  
,
2
2
x
1  cos x
cos  
,
2
2
x
1  cos x
tan  
2
1  cos x
sin x
1  cos x


1  cos x
sin x
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Slide 6- 16
Example

Find sin ( /8) exactly.
Solution:

sin 4  
2

1  cos

4
2
2
2
1
2
2 2

2
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Slide 6- 17
Another Example

x
Simplify tan tan x  1.
2
x
1  cos x sin x
tan
tan
x

1


1
Solution:
2
sin x cos x
sin x(1  cos x)

1
sin x cos x
1  cos x

1
cos x
1
cos x


1
cos x cos x
 sec x  1  1
 sec x
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Slide 6- 18
6.3
Proving
Trigonometric Identities

Prove identities using other identities.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Logic of Proving Identities

Method 1: Start with either the left or the right side of
the equation and obtain the other side.

Method 2: Work with each side separately until you
obtain the same expression.
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Slide 6- 20
Hints for Proving Identities






Use method 1 or 2.
Work with the more complex side first.
Carry out any algebraic manipulations, such as adding,
subtracting, multiplying, or factoring.
Multiplying by 1 can be helpful when rational
expressions are involved.
Converting all expressions to sines and cosines is often
helpful.
Try something! Put your pencil to work and get
involved. You will be amazed at how often this leads to
success.
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Slide 6- 21
Example

Prove the identity (tan 2 x  1)(cos 2 x  1)   tan 2 x .
Solution: Start with the left side.
(tan 2 x  1)(cos 2 x  1)   tan 2 x
 sin 2 x 
2
2
x
tan


1)

x
(cos
1



2
 cos x 
sin 2 x
2
2
2
x
tan


1

x
cos

sin x 
2
cos x
sin 2 x
2
2
2
x
tan


1

sin x  cos x 
2
cos x
sin 2 x
2
1

1


tan
x
2
cos x
sin 2 x
2



tan
x
2
cos x
 tan 2 x   tan 2 x
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Slide 6- 22
Another Example

Prove the identity:
1
 csc x  sin x
sec x tan x
Solution: Start with the
right side.
1
 csc x  sin x
sec x tan x
1

 sin x
sin x
1
sin 2 x


sin x sin x

Solution continued
1
1  sin 2 x

sec x tan x
sin x
cos 2 x

sin x
cos x cos x


sin x 1
 cot x  cos x
1
1


tan x sec x
1
1

sec x tan x sec x tan x
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Slide 6- 23
Example

tan x  cot y
 tan y  cot x.
Prove the identity
tan x cot y

Solution: Start with the left side.
tan x  cot y
 tan y  cot x
tan x cot y
tan x
cot y


tan x cot y tan x cot y
1
1


cot y tan x
tan y  cot x  tan y  cot x
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Slide 6- 24
6.4
Inverses of the Trigonometric
Functions

Find values of the inverse trigonometric functions.

Simplify expressions such as sin (sin–1 x) and
sin–1 (sin x).

Simplify expressions involving compositions such as
sin (cos–1 21 ) without using a calculator.

Simplify expressions such as sin arctan (a/b) by
making a drawing and reading off appropriate ratios.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Inverse Trigonometric Functions
Function
Domain
Range
y  sin 1 x
 arcsin x, where x  sin y
[1, 1]
[ / 2,  / 2]
y  cos 1 x
 arccos x, where x  cos y
[1, 1]
[0, ]
y  tan 1 x
 arctan x, where x  tan y
(, )
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  / 2, / 2 
Slide 6- 26
Example

Find each of the
following:
a) sin 1
Solution: a)
 Find  such that
3
sin  
2
3
2

3
1
b) cos  

 2 


c) tan 1 (1)
 would represent a 60°
or 120° angle.
3
sin
 60 and 120
2

2
or
and
3
3
1
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Slide 6- 27
Solution continued
b) Find  such that
 3
cos 
2



 would represent a 30°
reference angle in the 2nd and
3rd quadrants.
Therefore,  = 150° or 210°

3
cos  
  150 and 210
 2 
5
7
or
and
6
6
1
c) Find  such that
tan   1.



This means that the sine and
cosine of  must be
opposites.
Therefore,  must be 135°
and 315°.
tan 1 (1)  135 and 315
3
7
or
and
4
4
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Slide 6- 28
Domains and Ranges
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Slide 6- 29
Composition of Trigonometric Functions
sin(sin 1 x)  x,
for all x in the domain of sin 1 .
cos(cos 1 x)  x,
for all x in the domain of cos 1 .
tan(tan 1 x)  x,
for all x in the domain of tan 1.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6- 30
Examples

Simplify:

 1  1  
sin  sin    
 2 


Since 1/2 is in the
domain of sin–1,
 1  1  
1
sin  sin      
2
 2 

Simplify:
 1  3 2  
cos  cos 
 

 2 


3 2
Since
is not in the
2
domain of cos–1,
 1  3 2  
cos  cos 
 

 2 

does not exist.
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Slide 6- 31
Special Cases
sin 1 (sin x)  x,
for all x in the range of sin 1 .
cos 1 (cos x)  x,
for all x in the range of cos 1 .
tan 1 (tan x)  x,
for all x in the range of tan 1.
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Slide 6- 32
Examples

Simplify:

 
sin 1  sin 
2


Since /2 is in the range
of sin–1,
  
sin 1  sin  
2 2

Simplify:


tan  tan 
3

1

Since /3 is in the range
of tan–1,
 

tan 1  tan  
3 3

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Slide 6- 33
More Examples

Simplify:

 1 
sin  cos 1    
 2 

Solution:
2
 1
cos 1     120 or
3
 2
2
3
sin

3
2

Simplify:
  2  
tan 1  sin  

3

 
Solution:
2
3
sin

3
2

3
1
tan  
  40.9
 2 
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Slide 6- 34
6.5
Solving
Trigonometric Equations

Solve trigonometric equations.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Solving Trigonometric Equations

Trigonometric Equation—an equation that contains a
trigonometric expression with a variable.

To solve a trigonometric equation, find all values of the
variable that make the equation true.
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Slide 6- 36
Example


Solve 2 sin x  1 = 0.
Solution: First, solve for
sin x on the unit circle.
2sin x  1  0
2sin x  1

1
sin x 
2
6
1
x  sin 1  
 2
x  30 ,150 or
The values /6 and
5/6 plus any multiple
of 2 will satisfy the
equation. Thus the
solutions are
 5
 2 k and
5
 2 k
6
where k is any integer.
,
6 6
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Slide 6- 37
Graphical Solution

We can use either the Intersect method or the Zero
method to solve trigonometric equations. We graph the
equations y1 = 2 sin x  1 and y2 = 0.
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Slide 6- 38
Another Example

Solve 2 cos2 x  1 = 0.
Solution: First, solve for cos x on the unit circle.
x  45 ,135 , 225
 3 5
or ,
,
,
4 4
4
 3
2cos 2 x  1  0
2cos 2 x  1
1
2
cos x 
2
cos x  
The values
1
2
2
cos x  
2
,
,
,315
7
4
5 7
,
4 4
4
4
any multiple of 2 will satisfy
the equation.
plus
The solution can be written
as

4


2
k where k is any integer.
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Slide 6- 39
Graphical Solution


Solve 2 cos2 x  1 = 0.
One graphical solution shown.
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Slide 6- 40
One More Example

Solve 2 cos x + sec x = 0
Solution:
2cos x  sec x  0
1
2cos x 
0
cos x
1
2cos 2 x  1  0

cos x
1
0
cos x
1
0
cos x
or
Since neither factor of the equation
can equal zero, the equation has no
solution.

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
2cos 2 x  1  0
2cos 2 x  1
1
2
cos x 
2
1
cos x   
2
1
cos x  
2
Slide 6- 41
Graphical Solution

2 cos x + sec x
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Slide 6- 42
Last Example


Solve 2 sin2 x + 3sin x + 1 = 0.
Solution: First solve for sin x on the unit circle.
2sin x  3sin x  1  0
(2sin x  1)(sin x  1)  0
2
2sin x  1  0
2sin x  1
1
sin x  
2
1
1 
x  sin   
 2
7 11
x
,
6 6
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Slide 6- 43
Last Example continued

sin x  1  0
sin x  1

One Graphical Solution
1
x  sin ( 1)
3
x
2

x
7
11
3
 2 k ,
 2 k ,
 2 k
6
6
2
where k is any integer.
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Slide 6- 44