## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 85

#### Answer

The exact value of the expression is $\frac{\sqrt{3}}{2}$.

#### Work Step by Step

$\sin \left( -\frac{17\pi }{3} \right)$ lies in quadrant I. Add $6\pi$ to $-\frac{17\pi }{3}$ , to find a positive co-terminal angle less than $2\pi$. Consider $\alpha$ to be the positive co-terminal angle. \begin{align} & \alpha =\left( -\frac{17\pi }{3} \right)+6\pi \\ & =\frac{-17\pi +18\pi }{3} \\ & =\frac{\pi }{3} \end{align} The positive acute angle formed by the terminal side of $\theta$ and the x-axis is the reference angle ${\theta }'$. The reference angle of $-\frac{17\pi }{3}$ is $\frac{\pi }{3}$. The function value of the reference angle is $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ The angle $-\frac{17\pi }{3}$ is in quadrant I and the sine function is positive in quadrant I. Therefore, $\sin \left( -\frac{17\pi }{3} \right)=\sin \frac{\pi }{3}$ Substitute $\frac{\sqrt{3}}{2}$ for $\sin \frac{\pi }{3}$. $\sin \left( -\frac{17\pi }{3} \right)=\frac{\sqrt{3}}{2}$ The exact value of the expression is $\frac{\sqrt{3}}{2}$.

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