Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 6 - Review Questions - Page 501: 62


$sin(-\frac{5\pi}{4})=\frac{\sqrt 2}{2}$ $cos(-\frac{5\pi}{4})=-\frac{\sqrt 2}{2}$ $tan(-\frac{5\pi}{4})=-1$

Work Step by Step

Let $θ'$ be the reference angle. We must have that $0\leqθ'\lt2\pi$ $θ'=θ+n·2\pi$, where $n$ is an integer. $θ'=-\frac{5\pi}{4}+1·2\pi=\frac{3\pi}{4}$ $sin(-\frac{5\pi}{4})=sin\frac{3\pi}{4}=\frac{\sqrt 2}{2}$ $cos(-\frac{5\pi}{4})=cos\frac{3\pi}{4}=-\frac{\sqrt 2}{2}$ $tan(-\frac{5\pi}{4})=tan\frac{3\pi}{4}=-1$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.