Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 6 - The Circular Functions and Their Graphs - Test - Page 650: 9

Answer

$\frac{\sqrt 2}{2}$

Work Step by Step

As $\frac{3\pi}{4}$ is in quadrant II with a reference angle $\theta=\pi-\frac{3\pi}{4}=\frac{\pi}{4}$, we have $sin(\frac{3\pi}{4})=sin(\frac{\pi}{4})=\frac{\sqrt 2}{2}$
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