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 - Page 394: 22



Work Step by Step

$csc\alpha =csc\left( \alpha -2\pi n\right) \Rightarrow csc\dfrac {9\pi }{2}=csc\left( \dfrac {9\pi }{2}-4\pi \right) =csc\dfrac {\pi }{2}=\dfrac {1}{\sin \dfrac {\pi }{2}}=\dfrac {1}{1}=1$
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