This document discusses multiple angle identities in trigonometry. It covers double angle identities, power reducing identities, and half angle identities. Examples are provided to demonstrate how to use these identities to solve trigonometric equations and find unknown lengths in triangles. The document is a series of slides from a textbook on trigonometry focusing on important angle identities used in calculus.

1. 5.4
Multiple
Angle
Identities
Copyright © 2011 Pearson, Inc.

2. 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

3. Double Angle Identities
sin2u  2sinu cosu
cos2u 
cos2 u  sin2 u
2cos2 u 1
1 2sin2 u





tan2u 
2 tanu
1 tan2 u
Copyright © 2011 Pearson, Inc. Slide 5.4 - 3

4. Proving a Double-Angle Identity
cos2x  cos(x  x)
 cos x cos x - sin x sin x
 cos2 x  sin2 x
Copyright © 2011 Pearson, Inc. Slide 5.4 - 4

5. Power-Reducing Identities
sin2 u 
1 cos2u
2
cos2 u 
1 cos2u
2
tan2 u 
1 cos2u
1 cos2u
Copyright © 2011 Pearson, Inc. Slide 5.4 - 5

6. Example Reducing a Power of 4
Rewrite sin4 x in terms of trigonometric functions with
no power greater than 1.
Copyright © 2011 Pearson, Inc. Slide 5.4 - 6

7. Example Reducing a Power of 4
sin4 x  sin2 x2

1 cos2x
2

 

 
2

1 2cos2x  cos2 2x
4

1
4

cos2x
2

1
4
1 cos 4x
2

 

 

1
4

cos2x
2

1 cos 4x
8
Copyright © 2011 Pearson, Inc. Slide 5.4 - 7

8. Half-Angle Identities
sin
u
2
 
1 cosu
2
cos
u
2
 
1 cosu
2
tan
u
2


1 cosu
1 cosu
1 cosu
sinu
sinu
1 cosu









Copyright © 2011 Pearson, Inc. Slide 5.4 - 8

9. Example Using a Double Angle
Identity
Solve cos x  cos3x  0 in the interval [0,2 ).
Copyright © 2011 Pearson, Inc. Slide 5.4 - 9

10. Example Using a Double Angle
Identity
Solve cos x  cos3x  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

11. Example Using a Double Angle
Identity
Solve cos x  cos3x  0 in the interval [0,2 ).
Confirm Algebraically
cos x  cos3x  0
cos x  cos x cos2x  sin xsin2x  0
cos x  cos x1 2sin2 x sin x2sin x cos x 0
cos x  cos x  2cos x sin2 x  2cos x sin2 x  0
2cos x  4 cos x sin2 x  0
2cos x 1 2sin2  x 0
Copyright © 2011 Pearson, Inc. Slide 5.4 - 11

12. Example Using a Double Angle
Identity
2cos x1 2sin2 x 0
cos x  0 or 1 2sin2 x  0
x 

2
or
3
2
or sin x  
2
2
x 

2
or
3
2
or x 

4
,
3
4
,
5
4
, or
7
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

13. Quick Review
Find the general solution of the equation.
1. cot x 1  0
2. (sin x)(1 cos x)  0
3. cos x  sin x  0
4. 2sin x  22sin x 1 0
5. Find the height of the isosceles triangle with
base length 6 and leg length 4.
Copyright © 2011 Pearson, Inc. Slide 5.4 - 13

14. Quick Review Solutions
Find the general solution of the equation.
1. cot x 1  0 x 
3
4
  n
2. (sin x)(1 cos x)  0 x   n
3. cos x  sin x  0 x 

4
  n
Copyright © 2011 Pearson, Inc. Slide 5.4 - 14

15. Quick Review Solutions
Find the general solution of the equation.
4. 2sin x  22sin x 1 0
x 
5
4
 2 n, x 
7
4
 2 n,
x 

6
 2 n, x 
5
6
 2 n
5. Find the height of the isosceles triangle with
base length 6 and leg length 4. 7
Copyright © 2011 Pearson, Inc. Slide 5.4 - 15