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: 44



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 $\tan{60^o} = \sqrt3$. Thus, $\cot^{-1}{\left(\frac{\sqrt3}{3}\right)}=60^o$
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