Download Slide 5.4

5.4
Multiple
Angle
Identities
Copyright © 2011 Pearson, Inc.
What you’ll learn about




Double-Angle Identities
Power-Reducing Identities
Half-Angle Identities
Solving Trigonometric Equations
… and why
These identities are useful in calculus courses.
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 2
Double Angle Identities
sin 2u  2sinu cosu
cos 2 u  sin 2 u

2
cos 2u  2 cos u  1
1  2sin 2 u

2 tanu
tan 2u 
2
1  tan u
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 3
Proving a Double-Angle Identity
cos 2x  cos(x  x)
 cos x cos x - sin x sin x
 cos 2 x  sin 2 x
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 4
Power-Reducing Identities
1  cos 2u
sin u 
2
1  cos 2u
2
cos u 
2
1  cos 2u
2
tan u 
1  cos 2u
2
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 5
Example Reducing a Power of 4
Rewrite sin 4 x in terms of trigonometric functions with
no power greater than 1.
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 6
Example Reducing a Power of 4

sin x  sin x
4
2

2
 1  cos 2x 



2
2
1  2 cos 2x  cos 2 2x

4
1 cos 2x 1  1  cos 4x 
 
 


4
2
4
2
1 cos 2x 1  cos 4x
 

4
2
8
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 7
Half-Angle Identities
u
1  cosu
sin  
2
2
u
1  cosu
cos  
2
2
Copyright © 2011 Pearson, Inc.
 1  cosu

1

cosu

u 1  cosu
tan  
2  sinu
 sinu
1  cosu

Slide 5.4 - 8
Example Using a Double Angle
Identity
Solve cos x  cos 3x  0 in the interval [0,2 ).
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 9
Example Using a Double Angle
Identity
Solve cos x  cos 3x  0 in the interval [0,2 ).
Solve Graphically
The graph suggest that
there are six solutions:
0.79, 1.57, 2.36,
3.93, 4.71, 5.50.
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 10
Example Using a Double Angle
Identity
Solve cos x  cos 3x  0 in the interval [0,2 ).
Confirm Algebraically
cos x  cos 3x  0
cos x  cos x cos 2x  sin x sin 2x  0


cos x  cos x 1  2sin 2 x  sin x 2sin x cos x   0
cos x  cos x  2 cos x sin 2 x  2 cos x sin 2 x  0
2 cos x  4 cos x sin 2 x  0


2 cos x 1  2sin 2 x  0
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 11
Example Using a Double Angle
Identity


2 cos x 1  2sin 2 x  0
cos x  0

3
x  or
2
2

3
x  or
2
2
or
or
or
1  2sin x  0
2
2
sin x  
2
 3 5
7
x  , , , or
4 4 4
4
The six exact solutions in the given interval are
 
3 5 3
7
, ,
,
,
, and
.
4 2 4 4 2
4
Copyright © 2011 Pearson, Inc.
Slide 5.4 - 12