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Chapter 11: Trigonometric Identities and
Equations
11.1 Trigonometric Identities
11.2 Addition and Subtraction Formulas
11.3 Double-Angle, Half-Angle, and Product-Sum
Formulas
11.4 Inverse Trigonometric Functions
11.5 Trigonometric Equations
11.3 Double-Angle, Half-Angle, and Product-Sum
Formulas
Double-Angle Identities
E.g. cos 2A = cos(A + A)
= cos A cos A – sin A sin A
= cos² A – sin² A
Other forms for cos 2A are obtained by substituting either
cos² A = 1 – sin² A or sin² A = 1 – cos² A to get
cos 2A = 1 – 2 sin² A or cos 2A = 2 cos² A – 1.
cos 2 A  cos A  sin A
cos 2 A  1  2 sin A
cos 2 A  2 cos A  1
sin 2 A  2 sin A cos A
2
2
2
2
2 tan A
tan 2 A 
1 tan 2 A
11.3 Finding Function Values of 2
Example Given cos   53 and sin  < 0, find a) sin 2,
b) cos 2, and c) tan 2.
Solution To find sin 2, we must find sin .
sin 2   cos 2   1
2
a) sin 2  2 sin  cos 
 4  3 
 2   
 5  5 
24

25
3
sin      1
5
9
sin    1 
25
4 Choose the negative
sin   
square root since sin  < 0.
5
2
So cos 2 is…
8
5
94%
3%
a.
..
0%
he
3%
of
t
0%
No
ne
1.
4
2. 5
24
3. 7
7
4.  25
5. None of the above
11.3 Finding Function Values of 2
b) cos 2  cos 2   sin 2 
2
3  4
    
5  5
2
7

25
2 tan 
sin   54
4
c) tan 2 
,
where
tan





3
1  tan 2 
cos 
3
5

2 43 
1   43 
2
24

OR
7
sin 2
tan 2 
cos 2
 24
 257
 25
24

7
11.3 Simplifying Expressions Using Double-Number
Identities
Example Simplify each expression.
(a) cos² 7x – sin² 7x
(b) sin 15° cos 15°
Solution
(a) cos 2A = cos² A – sin² A. Substituting 7x in for A
gives cos² 7x – sin² 7x = cos 2(7x) = cos 14x.
(b) Apply sin 2A = 2 sin A cos A directly.
1
sin 15 cos15  (2) sin 15 cos15
2
1
 sin( 2 15 )  1 sin 30  1
2
2
4


11.3 Product-to-Sum Identities
• Product-to-sum identities are used in calculus to find
integrals of functions that are products of trigonometric
functions.
• Adding identities for cos(A + B) and cos(A – B) gives
cos( A  B)  cos A cos B  sin A sin B
cos( A  B)  cos A cos B  sin A sin B
cos( A  B)  cos( A  B)  2 cos A cos B
1
cos A cos B  [cos( A  B)  cos( A  B)].
2
11.3 Product-to-Sum Identities
• Similarly, subtracting and adding the sum and
difference identities of sine and cosine, we may
derive the identities in the following table.
Product-to-Sum Identities
cos A cos B  12 [cos( A  B)  cos( A  B)]
sin A sin B  12 [cos( A  B)  cos( A  B)]
sin A cos B  12 [sin( A  B)  sin( A  B)]
cos A sin B  12 [sin( A  B)  sin( A  B)]
11.3 Using a Product-to-Sum Identity
Example Rewrite cos 2 sin  as the sum or difference of
two functions.
Solution By the identity for cos A sin A, with 2 = A
and  = B,
cos 2 sin 
1
 [sin( 2   )  sin( 2   )]
2
1
1
 sin 3  sin  .
2
2
Since, cos A sin B  12 [sin( A  B)  sin( A  B)]
11.3 Sum-to-Product Identities
• From the previous identities, we can derive another
group of identities that are used to rewrite sums of
trigonometric functions as products.
Sum-to-Product Identities
sin A  sin B  2 sin  A2 B  cos  A2 B 
sin A  sin B  2 cos  A2 B sin  A2 B 
cos A  cos B  2 cos  A2 B  cos  A2 B 
cos A  cos B  2 sin  A2 B sin  A2 B 
11.3 Using a Sum-to-Product Identity
Example Write sin 2 – sin 4 as a product of two
functions.
Solution Use the identity for sin A – sin B, with 2 = A and
4 = B.
sin 2  sin 4
 2  4   2  4  Since, sin A  sin B  2 cos A B sin  AB 
2
2
 2 cos
 sin 

 2   2 
6
 2 
 2 cos sin   
2
 2 
 2 cos 3 sin(  )
 2 cos 3 sin 
Since sine is an odd function
11.3 Half-Number Identities
• Half-angle identities for sine and cosine are used in calculus when
eliminating the xy-term from an equation of the form Ax² + Bxy +
Cy² + Dx + Ey + F = 0, so the type of conic it represents can be
determined.
• From the alternative forms of the identity for cos 2A, we can
derive three additional identities, e.g. sin A .
2
cos 2x 1 2sin 2 x
2sin 2 x 1 cos 2x
1cos 2 x
A
sin x  
Let 2x  A so that x  .
2
2
1cos A Choose the sign ± depending on
A
sin  
the quadrant of the angle A/2.
2
2
11.3 Half-Number Identities
Half-Number Identities
A
1  cos A
cos  
2
2
A
1  cos A
sin  
2
2
A
1  cos A
A
sin A
tan  
tan 
2
1  cos A
2 1  cos A
A 1  cos A
tan 
2
sin A
11.3 Using a Half-Number Identity to Find an Exact Value

Example Find the exact value of cos .
12
Solution




3
6
cos
1 
2
 cos


12
2
2




22
1  cos
6

2
2 3

3
2
1
2

2
11.3 Finding Function Values of x/2
Example Given cos x  23 , with
3
2
 x  2 , find
cos 2x , sin 2x , and tan 2x .
Solution The half-angle terminates in quadrant II since
3
3
x

x

2



2
4
2  .
2
x
1

3 
a) sin

2
2
x
1  23
b) cos  

2
2
x
c) tan
2
sin

cos
x
2
x
2

1
6

6
6
5
30

6
6

6
6
30
6
5

5
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