Trigonometry 7th Edition

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

Chapter 1 - Section 1.4 - Introduction to Identities - 1.4 Problem Set - Page 41: 58

Answer

$\sin{\theta} =-0.41$ $\cos{\theta} =-0.91 $ $\tan{\theta} =0.45$ $\csc{\theta} = -2.45$ $\sec{\theta} =-1.09$ $\cot{\theta} =2.22 $

Work Step by Step

$\csc{\theta} = -2.45$ $\sin{\theta} = \dfrac{1}{\csc{\theta}} = \dfrac{1}{-2.45} \approx -0.41$ $\because \theta \in QIII \hspace{20pt} \therefore \cos{\theta}$ is negative. $\cos{\theta} = -\sqrt{1-\sin^2{\theta}} = - \sqrt{1-(-0.41)^2} \approx -0.91 $ $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} = \dfrac{-0.41}{-0.91} \approx 0.45$ $\sec{\theta} = \dfrac{1}{\cos{\theta}} = \dfrac{1}{-2.45} \approx -1.09$ $\cot{\theta} = \dfrac{\cos{\theta}}{\sin{\theta}} = \dfrac{-0.91}{-0.41} \approx 2.22 $
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.