Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding: 23

Answer

$\sqrt 2$

Work Step by Step

$\sec{\alpha} =\sec{\left( \alpha -2\pi k\right)} \\\Rightarrow \sec{\dfrac {17\pi }{4}}=\sec{\left( \dfrac {17\pi }{4}-4\pi \right)} \\=\sec{\dfrac {\pi }{4}} \\=\dfrac {1}{\cos \frac {\pi }{4}} \\=\dfrac {1}{\frac {\sqrt {2}}{2}} \\=\sqrt {2}$
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