Answer
(a) $[-\frac{\pi}{2}, \frac{\pi}{2}]$. $sin(y)=x$. $sin^{-1}\frac{1}{2}=\frac{\pi}{6}$, $sin(\frac{\pi}{6})=\frac{1}{2}$.
(b) $[0,\pi]$. $cos(y)=x$. $cos^{-1}\frac{1}{2}=\frac{\pi}{3}$, $cos\frac{\pi}{3}=\frac{1}{2}$
Work Step by Step
(a) restrict the domain of sine to the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
... $sin(y)=x$. For example, $sin^{-1}\frac{1}{2}=\frac{\pi}{6}$ because $sin(\frac{\pi}{6})=\frac{1}{2}$.
(b) restrict the domain of cosine to the interval $[0,\pi]$. ... $cos(y)=x$.
For example, $cos^{-1}\frac{1}{2}=\frac{\pi}{3}$ because $cos\frac{\pi}{3}=\frac{1}{2}$