Answer
(a) $\sin^{-1}\Big(\frac{-1}{2}\Big)=-\frac{\pi}{6}$
(b) $\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\frac{\pi}{4}$
(c) $\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)=-\frac{\pi}{3}$
Work Step by Step
*Recall the definition of arcsine:
$y=\sin^{-1}x$ is the number in $[-\pi/2,\pi/2]$ for which $\sin y=x$
(a) $$\sin^{-1}\Big(\frac{-1}{2}\Big)=a$$
According to the defintion: $$\sin a=-\frac{1}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[-\pi/2,\pi/2]$$
So $a=-\frac{\pi}{6}$. In other words, $$\sin^{-1}\Big(\frac{-1}{2}\Big)=-\frac{\pi}{6}$$
(b) $$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\sin^{-1}\Big(\frac{\sqrt2}{2}\Big)=a$$
According to the defintion: $$\sin a=\frac{\sqrt2}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[-\pi/2,\pi/2]$$
So $a=\frac{\pi}{4}$. In other words, $$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)=\frac{\pi}{4}$$
(c) $$\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)=a$$
According to the defintion: $$\sin a=\frac{-\sqrt3}{2}\hspace{1cm}\text{and}\hspace{1cm}a\in[-\pi/2,\pi/2]$$
So $a=-\frac{\pi}{3}$. In other words, $$\sin^{-1}\Big(\frac{-\sqrt3}{2}\Big)=-\frac{\pi}{3}$$
