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 118: 71

Answer

$s = \{-\frac{11\pi}{6}, -\frac{7\pi}{6}, -\frac{5\pi}{6}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}\}$

Work Step by Step

$3~tan^2~s = 1$ $tan^2~s = \frac{1}{3}$ $tan~s = \pm \frac{1}{\sqrt{3}}$ Since $\frac{y}{x} = \pm \frac{1}{\sqrt{3}}$, the angle $s$ makes an angle of $\frac{\pi}{6}$ with the x-axis. In the interval $[-2\pi, \pi)$: $s = \{-2\pi+\frac{\pi}{6}, -\pi-\frac{\pi}{6}, -\pi+\frac{\pi}{6}, -\frac{\pi}{6}, \frac{\pi}{6}, \pi-\frac{\pi}{6}\}$ $s = \{-\frac{11\pi}{6}, -\frac{7\pi}{6}, -\frac{5\pi}{6}, -\frac{\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}\}$
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