Chapter 6

Trigonometric Identities and Equations

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SECTION 6.3

Double-Angle and Half-Angle Identities OBJECTIVES 1 2 3

Use double-angle identities.

Use power-reducing identities.

Use half-angle identities.

DOUBLE-ANGLE IDENTITIES

sin 2

x

 2 sin

x

cos

x

cos 2

x

 cos 2

x

 sin 2

x

tan 2

x

 2 tan 1  tan 2

x x

cos 2

x

 1  2 sin 2

x

cos 2

x

 2 cos 2

x

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3

EXAMPLE 1 Using Double-Angle Identities  3  is in quadrant II, find the 5 exact value of each expression.

a. sin 2  b. cos2  c. tan 2  Solution Use identities to find sin

θ

and tan

θ

.

sin tan     sin  cos   2   4 / 5  3 / 5 1    9 4 25 3  4 5

θ

is in QII so sin > 0.

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4

EXAMPLE 1 Using Double-Angle Identities Solution continued a . sin 2   2 sin  cos   3 5   24 25 b. cos 2   cos 2   s in 2  2   3 5    9 25 7  25 16 25  5 2 © 2011 Pearson Education, Inc. All rights reserved

5

EXAMPLE 1 Using Double-Angle Identities Solution continued c. tan 2   1 2  ta n t a n  2    8 1  3 16 9   8  3 7 9  2     1      4 3 4 3       2 8 3      9 7    24 7 © 2011 Pearson Education, Inc. All rights reserved

6

EXAMPLE 3 Finding a Triple-Angle Identity for Sines Verify the identity sin 3

x

= 3 sin

x

– 4 sin 3

x

.

Solution sin 3

x

= sin (2

x

+

x

) = sin 2

x

cos

x

+ cos 2

x

sin

x

= ( 2 sin

x

cos

x

) cos

x

+ ( 1 – 2 sin 2

x

) sin

x

= 2 sin

x

cos 2

x

+ sin

x

– 2 sin 3

x

= 2 sin

x

( 1 – sin 2

x

) + sin

x

– 2 sin 3

x

= 2 sin

x

– 2 sin 3

x

+ sin

x

– 2 sin 3

x

= 3 sin

x

– 4 sin 3

x

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7

POWER REDUCING IDENTITIES

sin 2

x

 1  cos 2

x

2 cos 2

x

 1  cos 2

x

2 tan 2

x

 1  1  cos 2

x

cos 2

x

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8

EXAMPLE 4 Using Power-Reducing Identities Write an equivalent expression for cos 4

x

that contains only first powers of cosines of multiple angles.

Solution Use power-reducing identities repeatedly.

cos 4

x

  cos 2

x

 2    1  cos 2

x

2   2  1 4 

x

 2 cos 2

x

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9

EXAMPLE 4 Using Power-Reducing Identities Solution continued  1 4   1  2 cos 2

x

 1  cos 4

x

2    1 4   1  2 cos 2

x

 1 2  1 2 cos 4

x

   1 4  2 4 cos 2

x

 1 8  1 8 cos 4

x

 3  8 1 2 cos 2

x

 1 8 cos 4

x

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sin  2  

HALF-ANGLE IDENTITIES

2  cos  2   2  tan  2     The sign, + or –, depends on the quadrant in  which lies.

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11

EXAMPLE 6 Using Half-Angle Identities Use a half-angle formula to find the exact value of cos 157.5º.

Solution 315  Because 157.5º = , use the half-angle identity  for cos with θ = 315°. Because 2 2   2 lies in quadrant II, cos is negative. 2  157 .

5  cos157.5º  cos 315º 2   2 © 2011 Pearson Education, Inc. All rights reserved

12

EXAMPLE 6 Using Half-Angle Identities Solution continued   1  2 4 º      1  2 2     2    2  2    2  2 2 © 2011 Pearson Education, Inc. All rights reserved

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