Trigonometry (10th Edition)

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

Chapter 3 - Radian Measure and the Unit Circle - Section 3.1 Radian Measure - 3.1 Exercises - Page 99: 82

Answer

$\dfrac{\sqrt3}{3}$

Work Step by Step

Convert the angle measure to degrees to obtain: $=-\frac{2\pi}{3} \cdot \frac{180^o}{\pi} = -2(60^o)=-120^o$ Thus, $\cot{(-\frac{2\pi}{3})} = \cot{(-120^o)}$ $-120^o$ is co-terminal with $-120^o+360^o=240^o$. $240^o$ is in Quadrant III so its reference angle is $=240^o-180^o=60^o$. Note that the cotangent function is positive in Quadrant III. From Section 2.1 (page 50) , we learned that: $\cot{60^o} = \dfrac{\sqrt3}{3}$ This means that: $\cot{(-\frac{2\pi}{3})} \\=\cot{(-120^o)} \\=\cot{60^o} \\= \dfrac{\sqrt3}{3}$
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