Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 497: 87

Answer

See below.

Work Step by Step

$LHS=\frac{(2cos^2\theta-1)^2}{cos^4\theta-sin^4\theta}=\frac{(cos^2\theta-sin^2\theta)^2}{(cos^2\theta+sin^2\theta)(cos^2\theta-sin^2\theta)}=\frac{cos^2\theta-sin^2\theta}{cos^2\theta+sin^2\theta}=cos^2\theta-sin^2\theta=1-2sin^2\theta=RHS$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.