Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 43



Work Step by Step

The value of $\theta$ must be in the interval $(0^o, 180^o) $. $\theta=\cot^{-1}{\left(-\frac{\sqrt3}{3}\right)}$ means that $\cot{\theta} = -\frac{\sqrt3}{3}$. Since cotangent is the reciprocal of tangent, then $\cot{\theta} = -\frac{\sqrt3}{3}$ means $\tan{\theta}=-\frac{3}{\sqrt3}=-\sqrt3$. Note that $\tan{120^o} = -\sqrt3$. Thus, $\cot^{-1}{\left(-\frac{\sqrt3}{3}\right)}=120^o$
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.