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 $sin(-3\pi)=sin (-\pi+2\pi)=sin (-\pi)=sin(-\pi+2\pi)=sin(\pi)=0$
Precalculus (10th Edition)
by Sullivan, Michael
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
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
