Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition
Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition
10th Edition
ISBN: 9780134673301
Author: Waits, Foley, Kennedy, Bock Demana
Publisher: PEARSON
Expert Solution & Answer
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Chapter 5.2, Problem 76E
Solution

To prove: The identity cos6x−sin6x=cos2x−sin2x1−cos2xsin2x .

Given information:

The given identity is cos6x−sin6x=cos2x−sin2x1−cos2xsin2x .

Formula used:

The trigonometric identity is sin2x+cos2x=1 .

The algebraic identity a3−b3=a−ba2+b2+ab .

Proof:

Simplify the left-hand side of the given equation using the algebraic identity as follows.

  cos6x−sin6x=cos2x3−sin2x3=cos2x−sin2xcos2x2+sin2x2+cos2xsin2x=cos2x−sin2xcos2x2+sin2x2+2cos2xsin2x−cos2xsin2x=cos2x−sin2xcos2x+sin2x2−cos2xsin2x

Again simplify the obtained expression using the trigonometric and algebraic identities as follows.

  cos6x−sin6x=cos2x−sin2x12−cos2xsin2x=cos2x−sin2x1−cos2xsin2x

Here, the left-hand side is equal to the right-hand side.

Hence, it is proved that cos6x−sin6x=cos2x−sin2x1−cos2xsin2x .

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Chapter 5.2, Problem 75E
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Chapter 5.2, Problem 77E

Chapter 5 Solutions

Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition

Ch. 5.1 - Prob. 1QRCh. 5.1 - Prob. 2QRCh. 5.1 - Prob. 3QRCh. 5.1 - Prob. 4QRCh. 5.1 - Prob. 5QRCh. 5.1 - Prob. 6QRCh. 5.1 - Prob. 7QRCh. 5.1 - Prob. 8QRCh. 5.1 - Prob. 9QRCh. 5.1 - Prob. 10QR
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