Precalculus (6th Edition) Blitzer

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

Chapter 4 - Mid-Chapter Check Point - Page 577: 19


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

Work Step by Step

The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps: a) Quadrant- I: $\theta $ b) Quadrant- II: $180^{\circ}-\theta $ c) Quadrant -III: $\theta - 180^o $ d) Quadrant -IV: $360^{\circ}-\theta $ The reference angle is equal to $ 2\pi- \dfrac{11\pi}{6}=\dfrac{\pi}{6}$ Since, $\sec (\dfrac{\pi}{6})=\dfrac{2 \sqrt 3}{3}$ So, $\sec \dfrac{11\pi}{6}=\dfrac{2 \sqrt 3}{3}$ because $\theta $ lies in Quadrant-IV.
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.