Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 3 - Section 3.2 - Radians and Degrees - 3.2 Problem Set - Page 134: 101

Answer

$\csc \theta=2$ $\cos \theta= -\frac{\sqrt 3}{2}$ $\sec \theta=-\frac{2}{\sqrt 3}$ $\tan \theta=-\frac{1}{\sqrt 3}$ $\cot\theta=-\sqrt 3$

Work Step by Step

Since $\sin \theta=\frac{1}{2}$, we have $\csc \theta=\frac{1}{\sin \theta}=2$ Now $\sin^{2}\theta+\cos^{2}\theta=1$, i.e., $\cos^{2}\theta=1-\sin^{2}\theta$ or $\cos^{2}\theta= 1-\frac{1}{4}=\frac{3}{4}$ Since $\theta$ terminates in QII, $\cos\theta$ is negative. $\cos \theta= -\frac{\sqrt 3}{2}$ $\sec \theta=\frac{1}{\cos\theta}=-\frac{2}{\sqrt 3}$ Further, we have $\tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{1}{2}}{-\frac{\sqrt 3}{2}}=-\frac{1}{\sqrt 3}$ and $\cot\theta=\frac{1}{\tan\theta}=-\sqrt 3$
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.