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 - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 200: 18



Work Step by Step

$$\tan\theta=-\frac{\sqrt7}{2}\hspace{1cm}\sec\theta\gt0$$ From a Pythagorean Identity $$\tan^2\theta+1=\sec^2\theta$$ combined with a Reciprocal Identity $$\sec\theta=\frac{1}{\cos\theta}$$ we have $$\tan^2\theta+1=\frac{1}{\cos^2\theta}$$ $$\frac{1}{\cos^2\theta}=(-\frac{\sqrt 7}{2})^2+1=\frac{7}{4}+1=\frac{11}{4}$$ $$\cos^2\theta=\frac{4}{11}$$ $$\cos\theta=\pm\frac{2}{\sqrt{11}}=\pm\frac{2\sqrt{11}}{11}$$ As shown above, $\sec\theta=\frac{1}{\cos\theta}$. This means the signs of $\sec\theta$ and $\cos\theta$ are the same. Therefore, since $\sec\theta\gt0$, $\cos\theta\gt0$. $$\cos\theta=\frac{2\sqrt{11}}{11}$$ Also, from Quotient Identities, $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ which means $$\sin\theta=\tan\theta\times\cos\theta$$ $$\sin\theta=(-\frac{\sqrt7}{2})\times(\frac{2\sqrt{11}}{11})=-\frac{\sqrt{77}}{11}$$
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.