Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 53

Answer

$-0.73$

Work Step by Step

Let $\theta = \csc^{-1} {\left(-\frac{3}{2}\right)}$ Then, $\csc{\theta} = -\dfrac{3}{2}$. Since $\sin{\theta}= \dfrac{1}{\csc{\theta}}$, then $\sin{\theta} = -\dfrac{2}{3}$ Thus, $\theta = \sin^{-1} {\left(-\frac{2}{3}\right)}\approx -0.73$ Hence, $\csc^{-1} {\left(-\frac{3}{2}\right)}= \sin^{-1} {\left(-\frac{2}{3}\right)} \approx \boxed{-0.73}$
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