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 - Page 547: 13


$\frac{2\sqrt 3}{3}$

Work Step by Step

($\frac{\sqrt 3}{2}$ , -$\frac{1}{2}$) sec $\frac{11\pi}{6}$ is computed by $\frac{1}{x}$. So sec $\frac{11\pi}{6}$ = $\frac{1}{\frac{\sqrt 3}{2}}$ = $\frac{2\sqrt 3}{3}$.
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