Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 30: 27



Work Step by Step

$\csc^{2}\theta= 1+\cot^{2}\theta=1+\frac{16}{9}=\frac{25}{9}$ $\sin^{2}\theta=\frac{1}{\csc^{2}\theta}=\frac{9}{25}$ $\cos^{2}\theta=1-\sin^{2}\theta=1-\frac{9}{25}=\frac{16}{25}$ $\implies \cos\theta=±\frac{4}{5}$ As $\cot \theta=\frac{4}{3}$, $\cot\theta>0$ $\implies \frac{\cos\theta}{\sin\theta}>0$ But $\sin\theta<0$ (given) This implies that $\cos\theta$ is also less than 0. Therefore, $\cos\theta=-\frac{4}{5}$
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.