## Precalculus (10th Edition)

Published by Pearson

# Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 59

#### Answer

$-\frac{\sqrt3}{2}$

#### Work Step by Step

We know that $\sin$, $\csc$, $\tan$ are odd functions, which means $f(-\theta)=-f(\theta).$ We also know that $\cos$, $\sec$, are even functions, which means $f(-\theta)=f(\theta).$ Therefore $\sin{(-60^{o})}=-\sin{(60^{o})}=-\frac{\sqrt3}{2}.$

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