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: 19


$$\tan4\theta=\frac{2\tan2\theta}{2-\sec^22\theta}$$ The proof is shown below.

Work Step by Step

$$\tan4\theta=\frac{2\tan2\theta}{2-\sec^22\theta}$$ We start from the right side. $$X=\frac{2\tan2\theta}{2-\sec^22\theta}$$ - Recall that $\sec^2x=\tan^2x+1$. So if we replace $x$ with $2\theta$, we end up with $\sec^22\theta=\tan^22\theta+1$. Thus, $$X=\frac{2\tan2\theta}{2-(\tan^22\theta+1)}$$ $$X=\frac{2\tan2\theta}{2-\tan^22\theta-1}$$ $$X=\frac{2\tan2\theta}{1-\tan^22\theta}$$ - Now remind yourself that $$\tan2x=\frac{2\tan x}{1-\tan^2x}$$ So again, if we replace $x$ with $2\theta$, the result would be $$\tan(2\times2\theta)=\frac{2\tan2\theta}{1-\tan^22\theta}$$ $$\tan4\theta=\frac{2\tan2\theta}{1-\tan^22\theta}$$ Therefore, $$X=\tan4\theta$$ which means $$\tan4\theta=\frac{2\tan2\theta}{2-\sec^22\theta}$$ and 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.