Trigonometry (11th Edition) Clone

$$y=-\frac{\pi}{2}$$
$$y=\sin^{-1}(-1)$$ First, we see that the domain of inverse sine function is $[-1,1]$. $-1$ lies in this range, so $\sin^{-1}(-1)$ 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 $y=\sin^{-1}(-1)$ into $\sin y=-1$ Normally, we can say $\sin\frac{3\pi}{2}=-1$ or $\sin(-\frac{\pi}{2})=-1$ (not to mention an infinite other results of $y$ that could satisfy $\sin y=-1$). However, there can be only one result of $y$ fitting in the given range $y\in[-\frac{\pi}{2},\frac{\pi}{2}]$, which is $-\frac{\pi}{2}$. Therefore, $$y=-\frac{\pi}{2}$$