Trigonometry (11th Edition) Clone

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

Chapter 6 - Review Exercises - Page 291: 12



Work Step by Step

RECALL: $y=\text{arccsc}{(x)} \longrightarrow \sin{y}=\frac{1}{x}$, $y$ is in the interval $[-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$ Thus, $y=\text{arccsc}{(\frac{2\sqrt3}{3})}$ implies that $\sin{y}=\frac{1}{\frac{2\sqrt3}{3}}=\frac{\sqrt3}{2}$. Note that $\sin{(\frac{\pi}{3})}=\frac{\sqrt3}{2}$. Therefore, $y=\text{arccsc}(\frac{2\sqrt3}{3})\longrightarrow y=\frac{\pi}{3}$
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