Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Chapter 7 Test Prep - Review Exercises - Page 737: 50


$2\cos^3 x-\cos x=\frac{\cos^2 x-\sin^2 x}{\sec x}$

Work Step by Step

Start with the right side: $\frac{\cos^2 x-\sin^2 x}{\sec x}$ Rewrite in terms of sine and cosine: $=\frac{\cos^2 x-\sin^2 x}{\frac{1}{\cos x}}$ Multiply top and bottom by $\cos x$: $=\frac{\cos^2 x-\sin^2 x}{\frac{1}{\cos x}}*\frac{\cos x}{\cos x}$ $=\cos x*(\cos^2 x-\sin^2 x)$ Rewrite $\sin^2 x$ as $1-\cos^2 x$: $=\cos x*(\cos^2 x-(1-\cos^2 x))$ Simplify: $=\cos x*(\cos^2 x-1+\cos^2 x)$ $=\cos x*(2\cos^2 x-1)$ $=2\cos^3 x-\cos x$ Since this equals the left side, the identity has been proven.
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.