Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Review Exercises - Page 646: 99


The exact value of the provided expression is $-\frac{\pi }{6}$.

Work Step by Step

Let us assume $\theta ={{\tan }^{-1}}\left( -\frac{\sqrt{3}}{3} \right)$. Then, we have $\tan \theta =\left( -\frac{\sqrt{3}}{3} \right)$ We know that for the tan function, the interval for the angle is $\left( -\frac{\pi }{2},\frac{\pi }{2} \right)$. Therefore, the only angle that satisfies $\tan \theta =\left( -\frac{\sqrt{3}}{3} \right)$ is $\left( -\frac{\pi }{6} \right)$. Hence, $\theta =\left( -\frac{\pi }{6} \right)$ and the exact value of the given expression is ${{\tan }^{-1}}\left( -\frac{\sqrt{3}}{3} \right)=-\frac{\pi }{6}$
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.