Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Summary Exercises on Verifying Trigonometric Identities - Page 245: 27


$$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$ The process is explained step by step below.

Work Step by Step

$$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$ Let's see to the left side first. $$X=\frac{2(\sin x-\sin^3x)}{\cos x}$$ $$X=\frac{2\sin x(1-\sin^2x)}{\cos x}$$ - As to $(1-\sin^2x)$, recall that $\cos^2x=1-\sin^2x$, meaning that $$X=\frac{2\sin x\cos^2x}{\cos x}$$ $$X=2\sin x\cos x$$ - Now recall that $2\sin x\cos x=\sin 2x$, as stated in double-angle identity for sine. Therefore, $$X=\sin2x$$ That means $$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$ The verification is complete. The equation is an identity.
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