Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 53



Work Step by Step

$\dfrac{17\pi}{6}$ is close to $\dfrac{18\pi}{6}=3\pi$. This means that this angle is between $2.5\pi$ and $3\pi$. RECALL: An angle $\theta$, where $2.5\pi \lt \theta \lt 3\pi$, is coterminal with: $\theta - 2\pi$ Thus, the given angle is coterminal with: $=\dfrac{17\pi}{6} - 2\pi = \dfrac{17\pi}{6} - \dfrac{12\pi}{6} = \dfrac{5\pi}{6}$ $\dfrac{17\pi}{6}$ is coterminal with $\dfrac{5\pi}{6}$. $\dfrac{5\pi}{6}$ is in Quadrant II so $\dfrac{17\pi}{6}$ is also in Quadrant II. RECALL: The following are the means on how to find the reference angle of an angle $0 \leq \theta \lt 2\pi$ based on its position: (1) Quadrant I: $\theta$ (itself) (2) Quadrant II: $\pi-\theta$ (3) Quadrant III: $\theta - \pi$ (4) Quadrant IV: $2\pi - \theta$ Use formula (2) above to obtain: reference angle of $\dfrac{17\pi}{6}$ = reference angle of $\dfrac{5\pi}{6}$, which is $\\=\pi - \dfrac{5\pi}{6} \\=\dfrac{6\pi}{6} - \dfrac{5\pi}{6} \\=\dfrac{\pi}{6}$
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