Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Skill Practice - Page 871: 31


$\dfrac{2\sqrt 3}{3}$

Work Step by Step

When $P$ is the period of $f(x)$ then we have $ f(x+p)=f(x)$ Here, we have $ \theta= \dfrac{11 \pi}{6}$ lies in the fourth quadrant thus, the angle is in the interval $[0, 2 \pi]$. But $\theta'=2 \pi-\dfrac{11 \pi}{6}=\dfrac{\pi}{6}$ and $\sec \theta'=\sec \dfrac{\pi}{6}=\dfrac{1}{\cos \dfrac{\pi}{6}}$ $\dfrac{1}{\cos \dfrac{\pi}{6}}=\dfrac{1}{\sqrt 3/2}$ or, $=\dfrac{2}{\sqrt 3}$ or, $=\dfrac{2\sqrt 3}{3}$
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