Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Review - Exercises - Page 529: 39

Answer

$-\dfrac {\sqrt {2}}{2} $

Work Step by Step

Period of $cos$ function is $2\pi$ It means $\cos \alpha =\cos \left( 2\pi +\alpha \right) =\cos \left( -2\pi +\alpha \right) $ $\cos \left( 585\right) =\cos \left( 360+225\right) =\cos \left( 2\pi +\dfrac {5\pi }{4}\right) =\cos (\dfrac {5\pi }{4})=\cos \left( \pi +\dfrac {\pi }{4}\right) =\cos \pi \cos \dfrac {\pi }{4}-\sin \pi \sin \dfrac {\pi }{4}=\left( -1\right) \times \cos \dfrac {\pi }{4}-0=-\cos \dfrac {\pi }{4}=-\dfrac {\sqrt {2}}{2} $
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