Trigonometry (11th Edition) Clone

$$y=0$$
$$y=\sin^{-1}0$$ First, we see that the domain of inverse sine function is $[-1,1]$. $0$ lies in this range, so $\sin^{-1}0$ does exist. Also, it should be noted that the range of inverse sine function is $[-\frac{\pi}{2},\frac{\pi}{2}]$. In other words, $y\in[-\frac{\pi}{2},\frac{\pi}{2}]$. We can rewrite $y=\sin^{-1}0$ as $\sin y=0$ We see that $\sin 0=0$. Since $0\in[-\frac{\pi}{2},\frac{\pi}{2}]$, we can conclude $y=0$