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: 85

Answer

$\sqrt3$

Work Step by Step

Convert the angle measure to degrees to obtain: $=-\frac{14\pi}{3} \cdot \frac{180^o}{\pi} = -14(60)^o=-840^o$ Thus, $\tan{(-\frac{14\pi}{3})} = \tan{(-840^o)}$ $-840^o$ is co-terminal with $-840^o+1080^o=240^o$. $240^o$ is in Quadrant III so its reference angle is $=240^o-180^o=60^o$. Note that the tangent function is positive in Quadrant III.
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