## Precalculus (6th Edition)

Published by Pearson

# Chapter 6 - The Circular Functions and Their Graphs - 6.2 The Unit Circle and Circular Functions - 6.2 Exercises - Page 589: 72

#### Answer

$\color{blue}{s=\dfrac{7\pi}{6}}$

#### Work Step by Step

Use the inverse sine function of a calculator in radian mode to obtain: $s=\sin^{-1}{(-\frac{1}{2})} = -\frac{\pi}{6}$ $-\dfrac{\pi}{6}$ is coterminal with $-\frac{\pi}{6}+2\pi=\frac{11\pi}{6}$. The angle must be in the given interval, which is actually in Quadrant III. Note that $\frac{11\pi}{6}$ is $\frac{\pi}{6}$ less than $2\pi$ (the positive x-axis). This angle will have the same sine value as the angle that is $\frac{\pi}{6}$ more than $\pi$ (the negative y-axis), which is the angle $\frac{7\pi}{6}$. Thus, $\sin{\frac{11\pi}{6}}=\sin{\frac{7\pi}{6}}$ Therefore. $\color{blue}{s=\dfrac{7\pi}{6}}$

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