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: 41

Answer

$-\dfrac {2}{\sqrt {3}} $

Work Step by Step

Period of $csc$ function is $2\pi$ It means $csc\left( \alpha \right) =csc\left( \alpha +2\pi k\right) =csc\left( \alpha -2\pi k\right) $ $csc\left( \dfrac {8\pi }{3}\right) =csc\left( 2\pi +\dfrac {2\pi }{3}\right) =csc\dfrac {2\pi }{3}=\dfrac {1}{\sin \dfrac {2\pi }{3}}==\dfrac {1}{\sin \left( \pi -\dfrac {\pi }{3}\right) }=\dfrac {1}{\sin \pi \cos \dfrac {\pi }{3}-\cos \pi \sin \dfrac {\pi }{3}}=\dfrac {1}{0-\left( -1\right) \times \sin \dfrac {\pi }{3}}=\dfrac {1}{-\sin \dfrac {\pi }{3}}=-\dfrac {1}{\dfrac {\sqrt {3}}{2}}=-\dfrac {2}{\sqrt {3}} $
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