Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Section 5.1 - The Unit Circle - 5.1 Exercises - Page 408: 52

Answer

(a) Reference Number: $\pi/6$ (b) Terminal Point: $(-\sqrt 3/2, -1/2)$

Work Step by Step

1. $31\pi/6 \gt 2\pi$. Subtract $2\pi$ from it until the result is less than $2\pi$. $$\frac{31\pi}{6} - 2\pi = -\frac{31\pi}{6} - \frac{12\pi}{6} = \frac{19\pi}{6}$$ $$\frac{19\pi}{6} - 2\pi = \frac{7\pi}{6}$$ 2. $7\pi/6$ is in Quadrant III, thus, the Reference Number is given by the equation: $$t^- = \frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi} 6$$ 3. According to the Table 1, the terminal point for $\pi/6$ is $(\sqrt 3/2, 1/2)$. In Quadrant III: x-coordinate is negative and y-coordinate is also negative. Therefore, the terminal point is: $(-\sqrt 3/2, -1/2)$
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