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: 81

Answer

$-\dfrac{\sqrt3}{2}$

Work Step by Step

Convert the angle measure to degrees to obtain: $=-\frac{8\pi}{3} \cdot \frac{180^o}{\pi} = -8(60^o)=-480^o$ Thus, $\sin{(-\frac{8\pi}{3})} = \sin{(-480^o)}$ $-480^o$ is co-terminal with $-480^o+720^o=240^o$. $240^o$ is in Quadrant III so its reference angle is $=240^o-180^o=60^o$. Note that the sine function is negative in Quadrant III. From Section 2.1 (page 50) , we learned that: $\sin{60^o} = \dfrac{\sqrt3}{2}$ This means that: $\sin{(-\frac{8\pi}{3})} \\=\sin{(-480^o)} \\=-\sin{60^o} \\= -\dfrac{\sqrt3}{2}$
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