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

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Unit 5.4

  • 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