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: 60


$sin(\frac{5\pi}{6})=\frac{1}{2}$ $cos(\frac{5\pi}{6})=-\frac{\sqrt 3}{2}$ $tan(\frac{5\pi}{6})=-\frac{\sqrt 3}{3}$

Work Step by Step

Let $θ'$ be the reference angle. We must have that $0\leqθ'\lt2\pi$ But, since $0\leqθ=\frac{5\pi}{6}\lt2\pi$, it is not necessary to find a reference angle. $sin(\frac{5\pi}{6})=\frac{1}{2}$ $cos(\frac{5\pi}{6})=-\frac{\sqrt 3}{2}$ $tan(\frac{5\pi}{6})=-\frac{\sqrt 3}{3}$
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.