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 3.3 The Unit Circle and Circular Functions - 3.3 Exercises - Page 117: 17

Answer

$\frac{\sqrt3}{2}$

Work Step by Step

RECALL: $\sin{s} = y \\\cos{s} = x \\\tan{s} = \frac{y}{x} \\\cot{s} = \frac{x}{y} \\\sec{s} = \frac{1}{x} \\\csc{s}=\frac{1}{y}$ (refer to Figure 11 , page 111 of the textbook) Moving $\frac{4\pi}{3}$ in counterclockwise (negative direction) yields the same terminal side as $\frac{2\pi}{3}$. This angle intersects the unit circle at the point $(-\frac{1}{2}, \frac{\sqrt3}{2})$. This point has: $x= -\frac{1}{2}$ $y=\frac{\sqrt3}{2}$ Thus, $\sin{(-\frac{4\pi}{3})} \\= \sin{\frac{2\pi}{3}} \\=\frac{\sqrt3}{2}$
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