Chapter 9, Problem 9CR

To determine

The exact value of the expression ${\sin }^{-1}\left(-\frac{1}{2}\right)$ in radian.

Answer to Problem 9CR

Solution:

The exact value of the expression ${\sin }^{-1}\left(-\frac{1}{2}\right)$ is $-\frac{\pi }{6}$ .

Explanation of Solution

Given information:

The expression ${\sin }^{-1}\left(-\frac{1}{2}\right)$

Let $\theta ={\sin }^{-1}\left(-\frac{1}{2}\right)$

Then $\theta$ is the angle, $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ , whose sine equals $-\frac{1}{2}$ .

  $\theta ={\sin }^{-1}\left(-\frac{1}{2}\right)$ ; $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

  $\Rightarrow \sin \theta =-\frac{1}{2}$ ; $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$

Using table,

  $\begin{array}{cccccccccc}\theta & -\frac{\pi }{2}& -\frac{\pi }{3}& -\frac{\pi }{4}& -\frac{\pi }{6}& 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}\\ \sin \theta & -1& -\frac{\sqrt{3}}{2}& -\frac{\sqrt{2}}{2}& -\frac{1}{2}& 0& \frac{1}{2}& \frac{\sqrt{2}}{2}& \frac{\sqrt{3}}{2}& 1\end{array}$

The only angle $\theta$ in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ whose $\sin \theta =-\frac{1}{2}$ is $-\frac{\pi }{6}$ .

  $\Rightarrow {\sin }^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi }{6}$ .

Therefore, the exact value of the expression ${\sin }^{-1}\left(-\frac{1}{2}\right)$ is $-\frac{\pi }{6}$ .