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: 16

Answer

$-1$

Work Step by Step

$\sec{\alpha} =\sec{\left( \alpha -360^o\right)} \Rightarrow \sec{540^o}=\sec{\left( 540^o-360^o\right)} =\sec{180^o}=\dfrac {1}{\cos {180^o}}=\dfrac {1}{\cos{\pi} }=\dfrac {1}{-1}=-1$
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