## Precalculus (6th Edition)

Published by Pearson

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

#### Answer

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

#### 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.

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