Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 258: 43

Answer

$120^o$

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, $\cot^{-1}{\left(-\frac{\sqrt3}{3}\right)}=120^o$
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