Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 417: 26



Work Step by Step

$\sec{(\theta+ 2 \pi k}) = \sec{\theta}$ $\therefore \sec{\dfrac{25}\pi{6}} = \sec{\left(\dfrac{\pi}{6}+ (2 \pi \times 2) \right)} = \sec{\dfrac{\pi}{6}}$ $\sec{\dfrac{\pi}{6}} \\\\ = \dfrac{1}{\cos{\frac{\pi}{6}}}\\\\ =\dfrac{1}{\frac{\sqrt3}{2}}\\\\ =1 \cdot \dfrac{2}{\sqrt3}\\\\ =\dfrac{2}{\sqrt3}\\\\ =\boxed{\dfrac{2\sqrt{3}}{3}}$
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