Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

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



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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