Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Skill Practice - Page 871: 27


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

Work Step by Step

When $P$ is the period of $f(x)$ then we have $ f(x+p)=f(x)$ Here, we have $ \csc (-420^{\circ})=-\csc 420^{\circ}=-\csc 60^{\circ} =-\dfrac{1}{\sin 60^{\circ}}$ and $-\dfrac{1}{\sin 60^{\circ}}=-\dfrac{1}{\sqrt 3/2}$ or, $=-\dfrac{2}{\sqrt 3}$ or, $=-\dfrac{2 \sqrt 3}{3}$
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