Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Summary Exercises on Verifying Trigonometric Identities - Page 239: 7

Answer

$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$ 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.
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.