Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Test - Page 293: 3d



Work Step by Step

RECALL : (1) For $y=\csc^{-1}{(x)}\longrightarrow \csc{y}=x,$ where $y$ must be in the interval $[-90^o, 0) \cup (0, 90^o]$. (2) $\csc{\theta} = \frac{1}{\sin{\theta}}$ (3) $y=\sin^{-1}{(x)} \longrightarrow \sin{y}=x$, where $y$ is in the interval $[-90^o, 90^o]$ Thus, $\theta=\csc^{-1}{\left(-\frac{2\sqrt3}{3}\right)}\longrightarrow \csc{\theta}=-\frac{2\sqrt3}{3}$ Using the definition in (2) above gives: \begin{array}{ccc} &\csc{\theta} &= &-\frac{2\sqrt3}{3} \\&\frac{1}{\sin{\theta}}&=&-\frac{2\sqrt3}{3} \\&1(3) &=&-2\sqrt3\cdot (\sin{\theta}) \\&3 &=&-2\sqrt3\cdot (\sin{\theta}) \\&\frac{3}{-2\sqrt3}&=&\sin{\theta} \\&-\frac{\sqrt3}{2}&=&\sin{\theta} \\&\theta&=&\sin^{-1}{(-\frac{\sqrt3}{2})} \end{array} Use a scientific calculator's inverse tangent function in degree mode to obtain: $\theta = \sin^{-1}{(-\frac{\sqrt3}{2})} \\\theta=-60^o$
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