Download PowerPoint for 9.3

9.3 Further Identities
• Double-Number 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.
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-1
9.3 Double-Number Identities
Double-Number Identities
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
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-2
9.3 Finding Function Values of 2
Example Given cos   53 and sin  < 0, find sin 2,
cos 2, and tan 2.
Solution To find sin 2, we must find sin .
sin 2   cos 2   1
2
3
4

sin      1  sin   
5
5
sin 2  2 sin  cos 
2
Choose the negative square
root since sin  < 0.
4  3 
24

sin 2  2     
25
 5  5 
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-3
9.3 Finding Function Values of 2
cos 2  cos   sin 
2
2
2
2
3  4
7

       
25
5  5
2 tan 
sin   54
4
tan 2 
, where tan 
 3 
2
1  tan 
cos 
3
5
2 43 
24


or
2
7
1   43 
sin 2  24
24
25
tan 2 
 7 
cos 2  25 7
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-4
9.3 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 cos 15  (2) sin 15 cos15
2
1
1
1


 sin( 2  15 )  sin 30 
2
2
4
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-5
9.3 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
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-6
9.3
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)]
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-7
9.3 Using a Product-to-Sum Identity
Example Rewrite cos 2 sin  as either the sum or
difference of two functions.
Solution By the identity for cos A sin A, with 2 = A
and  = B,
1
cos 2 sin   [sin( 2   )  sin( 2   )]
2
1
1
 sin 3  sin  .
2
2
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-8
9.3 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 
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-9
9.3 Using a Sum-to-Product Identity
Example Write sin 2t – sin 4t as a product of two
functions.
Solution Use the identity for sin A – sin B, with
2t = A and 4t = B.
 2t  4t   2t  4t 
sin 2t  sin 4t  2cos 
 sin 

 2   2 
6t
 2t 
 2cos sin   
2
 2
 2cos3t sin( t )
 2cos3t sin t
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-10
9.3 Half-Number Identities
• Half-number or 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 2 x  1  2 sin x
2
2 sin x  1  cos 2 x
2
A
1cos 2 x
sin x  
Let 2 x  A so that x  .
2
2
A
1cos A Choose the sign ± depending on
sin  
the quadrant of the angle A/2.
2
2
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-11
9.3
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
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-12
9.3 Using a Half-Number Identity to
Find an Exact Value
Example Find the exact value of cos
Solution
cos

12
.


 cos 6
12
2

1  cos

6
2
3

3
1 
2
1
2 3
2 

2



2
22
2
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-13
9.3 Finding Function Values of x/2
Example Given cos x  23 , with
cos 2x , sin 2x , and tan 2x .
3
2
 x  2 , find
Solution The half-angle terminates in quadrant II
since 32  x  2  34  2x   .
x
1  23
1
6
sin 


2
2
6 6
x
1  23
5
30
cos  


2
2
6
6
6
x sin 2x
5
6
tan 
 30  
x
2 cos 2  6
5
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-14
9.3 Simplifying Expressions Using
Half-Number Identities
1

cos
12
x
Example Simplify the expression 
.
2
Solution This matches the part of the identity for
cos A/2. Replace A with 12x to get
1  cos12 x
12 x

 cos
2
2
 cos 6 x.
Copyright © 2011 Pearson Education, Inc.
Slide 9.3-15