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: 7


$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$ The proof is below in the work step by step.

Work Step by Step

$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$ We examine the right side first. $$A=\frac{2\tan\theta}{1+\tan^2\theta}$$ As from quotient identity: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ so, $$A=\frac{\frac{2\sin\theta}{\cos\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$$ $$A=\frac{\frac{2\sin\theta}{\cos\theta}}{\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}}$$ Now recall that $\cos^2\theta+\sin^2\theta=1$. Thus, $$A=\frac{\frac{2\sin\theta}{\cos\theta}}{\frac{1}{\cos^2\theta}}$$ $$A=\frac{2\sin\theta\cos^2\theta}{\cos\theta\times1}$$ $$A=2\sin\theta\cos\theta$$ Finally, we can rewrite $2\sin\theta\cos\theta$ into $\sin2\theta$, according to double-angle identity for sines. $$A=\sin2\theta$$ Therefore, $$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$ The equation is an identity.
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