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Section 4.4
Trigonometric Expressions and Identities
In this section, you’ll learn to simplify trig expressions using identities and using basic algebraic
operations. You can add, subtract, multiply, divide and factor trig expressions, in much the same
manner than you can with algebraic expressions.
Sometimes, you can use trig identities to help you simplify trig expressions. Here is a list of trig
identities we have already met. Note that NONE of these identities will be provided on the tests.
You must know all of these identities.
sin(t )
cos(t )
cos(t )
cot(t ) 
sin(t )
tan(t ) 
Reciprocal Identities:
1
, sin(t )  0
sin(t )
1
sec(t ) 
, cos(t )  0
cos(t )
1
cot(t ) 
, tan(t )  0
tan(t )
csc(t ) 
Pythagorean Identities
sin 2 (t )  cos 2 (t )  1
1  tan 2 (t )  sec 2 (t )
1  cot 2 (t )  csc 2 (t )
You should know all three of the Pythagorean Identities or be able to derive the last two from the
first one.
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Opposite Angle Identities
sin(t )   sin(t )
cos(t )  cos(t )
tan(t )   tan(t )
csc(t )   csc(t )
sec(t )  sec(t )
cot(t )   cot(t )
We’ll use these in the next several examples.
Example 6: Simplify:
[1  cos( x)][csc( x)  cot( x)]
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Example 7: Simplify:
sin( x)
cos 2 ( x)  1
3
Example 8: Simplify:
sec 2  
tan    cot  
4
Example 9: Simplify:
1  sin 2 ( )
cos 2 ( )  1
5
Example 10: Simplify:
cot( x)
cot( x)

csc( x)  1 csc( x)  1
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You can also use the identities to help you solve problems like this one. (Note: you can also use
a triangle to help you work this problem.)
Example 11: If
cot( ) 
5
3
, where    
, find the exact values of tan( ) and sec( ) .
12
2
7
At times, you may be asked to verify identities. To do this, you’ll use the identities and algebraic
operations to show that the left-hand side of the problem equals the right-hand side of the
problem.
Here are some pointers for helping you verify identities:
1. Remember that your task is to show that the two sides of the equation are equal. You
may not assume that they are equal.
2. Choose one side of the problem to work with and leave the other one alone. You’ll use
identities and algebra to convert one side so that it is identical to the side you left alone.
You’ll work with the “ugly” or more complicated side.
3. If is often helpful to convert all trig functions into sine and cosine. This is usually very
helpful! (Unless it makes things worse!!)
4. Find common denominators, if appropriate.
5. Don’t try to do too much in one step. Take it one step at a time!
6. If working with one side doesn’t get you anywhere, try working with the other side
instead.
Example 12: Prove the identity:
sin x cos x
 cot x
1  cos 2 x
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EXERCISE: Prove the identity: cos 2 ( x)  sin 2 ( x) 
1  tan 2 ( x)
1  tan 2 ( x)
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Section 5.1
Trigonometric Functions of Real Numbers
Here are some identities you need to know:
sin(t )
cos(t )
cos(t )
cot(t ) 
sin(t )
tan(t ) 
Reciprocal Identities
1
, sin(t )  0
sin(t )
1
sec(t ) 
, cos(t )  0
cos(t )
1
cot(t ) 
, tan(t )  0
tan(t )
csc(t ) 
Opposite Angle Identities
sin(t )   sin(t )
cos(t )  cos(t )
tan(t )   tan(t )
csc(t )   csc(t )
sec(t )  sec(t )
cot(t )   cot(t )
Pythagorean Identities
sin 2 (t )  cos 2 (t )  1
1  tan 2 (t )  sec 2 (t )
1  cot 2 (t )  csc 2 (t )
Periodicity
sin(t  2k )  sin(t )
cos(t  2k )  cos(t )
sec(t  2k )  sec(t )
csc(t  2k )  csc(t )
tan(t  k )  tan(t )
cot(t  k )  cot(t )
(for all real numbers t and all integers k.)
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Example 1 : Simplify:
Example 2 : Simplify:
cot(t )
cos(t )
sin(t  6 )csc(t  2 )
cot(t   ) tan(t  5 )
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Example 3: Simplify:
EXERCISE: Simplify:
cos(t )  cos(t ) tan 2 (t )
sec(t  4 )  csc(t  6 )
1  tan(t  5 )
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Chapter 6
Section 6.1 - Sum and Difference Formulas
Note:
sin( A  B )  sin( A)  sin( B)
cos( A  B)  cos( A)  cos( B)
Sum and Difference Formulas for Sine, Cosine and Tangent
sin( A  B )  sin A cos B  sin B cos A
sin( A  B)  sin A cos B  sin B cos A
cos( A  B)  cos A cos B  sin A sin B
cos( A  B)  cos A cos B  sin A sin B
tan( A  B) 
tan A  tan B
1  tan A tan B
tan( A  B) 
tan A  tan B
1  tan A tan B
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3 

Example 2: Given that tan( x)  5 , evaluate tan x 
.
4 

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Example 3: Simplify each.
a. sin10 cos 55  sin 55 cos10
 
 7 
    7 
b. cos   cos 
  sin   sin 

 12 
 12 
 12   12 
c.
tan 40  tan 5
1  tan 40 tan 5
d.
tan 80  tan15
1  tan 80 tan15
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Example 4: Find the exact value of each. (Hint: use sum/difference formulas)
a. sin 15 0
 7 
b. cos 
 12 
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EXERCISE: Evaluate
 5 
tan  
 12 
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Example 5: Suppose that sin  
3
5

and cos  
where 0      . Find each of these:
5
13
2
a. sin(   )
b. cos(   )
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Example 6: Suppose cos  
1
7
and tan    where    ,   2 . Find
5
6
a. cos(   )
b. tan(   )
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EXERCISE: Suppose cos  
1
1
and cos    where 0     ,
2
5
    2 . Find
sin(   )
sin(   )
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Section 6.2 – Double and Half Angle Formulas
Now suppose we are interested in finding sin(2 A) . We can use the sum formula for sine to
develop this identity:
sin(2 A)  sin( A  A)
 sin A cos A  sin A cos A
 2 sin A cos A
Similarly, we can develop a formula for cos(2 A) :
cos(2 A)  cos( A  A)
 cos A cos A  sin A sin A
 cos 2 A  sin 2 A
We can restate this formula in terms of sine only or in terms of cosine only by using the
Pythagorean theorem and making a substitution. So we have:
cos(2 A)  cos 2  sin 2 A
 1  2 sin 2 A
 2 cos 2 A  1
We can also develop a formula for tan(2 A) :
tan(2 A)  tan( A  A)
tan A  tan A
1  tan A tan A
2 tan A

1  tan 2 A

These three formulas are called the double angle formulas for sine, cosine and tangent.
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Double – Angle Formulas
sin(2 A)  2 sin A cos A
cos(2 A)  cos 2 A  sin 2 A
tan(2 A) 
(Also: cos(2 A)  2 cos 2 A  1  1  2 sin 2 A )
2 tan A
1  tan 2 A
Now we’ll look at the types of problems we can solve using these identities.
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Example: Suppose that sin  
4
. Find
5
a) cos(2 )
b) sin(2 )
c) tan(2 )
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4

EXERCISE: Suppose that cos    and     . Find
2
7
a. cos(2 )
b. sin(2 )
c. tan(2 )
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Example 2: Simplify each:
a. 2sin 15  cos 15 
b. cos 2

5
 sin 2

5
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c.
2 tan15
1  tan 2 15
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Half – Angle Formulas
1  cos A
 A
sin    
2
2
1  cos A
 A
cos   
2
2
sin A
1  cos A
 A
tan  

sin A
 2  1  cos A
Note: In the half-angle formulas the symbol is intended to mean either positive or negative but
A
not both, and the sign before the radical is determined by the quadrant in which the angle
2
terminates.
Now we’ll look at the kinds of problems we can solve using half-angle formulas.
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Example 3: Use a half-angle formula to find the exact value of each.
a. sin 15
 5 
b. cos 

 8 
28
 7 
c. tan 
 12 
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Example 4: Answer these questions for cos  
4 3
   2 .
,
9 2
a. In which quadrant does the terminal side of the angle lie?
b. Complete the following: ___ 

2
 ___
c. In which quadrant does the terminal side of

2
lie?
 
d. Determine the sign of sin   .
2
 
e. Determine the sign of cos  .
2
 
f. Find the exact value of sin   .
2
 
g. Find the exact value of cos  .
2
 
h. Find the exact value of tan  .
2
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