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: 63


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

Work Step by Step

Simplify the left side: $\frac{\sin^2 x-\cos^2 x}{\csc x}$ Rewrite it in terms of sine and cosine: $=\frac{\sin^2 x-\cos^2 x}{\frac{1}{\sin x}}$ Multiply top and bottom by $\sin x$: $=\frac{\sin^2 x-\cos^2 x}{\frac{1}{\sin x}}*\frac{\sin x}{\sin x}$ $=(\sin^2 x-\cos^2x)\sin x$ $=-(\cos^2x-\sin^2 x)\sin x$ Use the identity $\cos^2x-\sin^2 x=\cos 2x$: $=-\cos 2x\sin x$ Simplify the right side: $2\sin^3 x-\sin x$ Factor out $\sin x$: $=\sin x(2\sin^2 x-1)$ $=-\sin x(1-2\sin^2 x)$ Use the identity $1-2\sin^2 x=\cos 2x$: $=-\sin x\cos 2x$ $=-\cos 2x\sin x$ Since the left side and the right side both simplify to $-\cos 2x\sin x$, they are equal to each other, and the identity has been proven.
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