Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set: 24

Answer

a.)-$\frac{\sqrt 3}{3}$ b.)$\frac{\sqrt 3}{3}$

Work Step by Step

a.)tan$\frac{11\pi}{6}$ corresponds to $\frac{y}{x}$ value. tan$\frac{11\pi}{6}$ = $\frac{\frac{-1}{2}}{\frac{\sqrt 3}{2}}$ = -$\frac{\sqrt 3}{3}$ b.)Tangent is an odd function, so tan(-t) = -tant. tan-$\frac{11\pi}{6}$ = - tan$\frac{11\pi}{6}$ = $\frac{\sqrt 3}{3}$
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.