## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 56

#### Answer

$\dfrac{\pi}{3}$

#### Work Step by Step

$\dfrac{17\pi}{3}$ is close to $\dfrac{18\pi}{3}=6\pi$. This means that this angle is between $5.5\pi$ and $6\pi$. RECALL: An angle $\theta$, where $5.5\pi \lt \theta \lt 6\pi$, is coterminal with: $\theta-4\pi$ Thus, the given angle is coterminal with: $=\dfrac{17\pi}{3}-4\pi=\dfrac{17\pi}{3} - \dfrac{12\pi}{3} = \dfrac{5\pi}{3}$ $\dfrac{5\pi}{3}$ is coterminal with $\dfrac{17\pi}{3}$. $\dfrac{5\pi}{3}$ is in Quadrant IV so $\dfrac{17\pi}{3}$ is also in Quadrant IV. 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 (4) above to obtain: reference angle of $\dfrac{17\pi}{3}$ = reference angle of $\dfrac{5\pi}{3}$, which is $\\2\pi - \dfrac{5\pi}{3} \\=\dfrac{6\pi}{3} - \dfrac{5\pi}{3} \\=\dfrac{\pi}{3}$

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