## Precalculus (10th Edition)

Published by Pearson

# Chapter 6 - Trigonometric Functions - 6.2 Trigonometric Functions: Unit Circle Approach - 6.2 Assess Your Understanding - Page 379: 28

#### Answer

$0$

#### Work Step by Step

We know that $sin$ has a period of $2\pi$, hence first we try and find a value where the argument is between $0$ and $2/pi$. Therefore $sin(-3\pi)=sin (-\pi+2\pi)=sin (-\pi)=sin(-\pi+2\pi)=sin(\pi)=0$

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