Trigonometry (11th Edition) Clone

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

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 264: 39



Work Step by Step

The value of $\theta$ must be in the interval $[-90^o, 90^o]$. Note that $\sin{60^o} = \frac{\sqrt{3}}{2}$. Since $\sin{-\theta} = -\sin{\theta}$, Then $\sin{(-60^o)}=-\sin{60^o} = -\frac{\sqrt{3}}{2}$ Thus, $\arcsin{(-\frac{\sqrt3}{2})}=-60^o$
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