Inverse Sine Function

Consider $\sin$ restricted to $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\to[-1,1]$. Graphically this restriction is one-one (strictly increasing) and onto $[-1,1]$, so its inverse exists, called the inverse sine function and denoted $\sin^{-1}:[-1,1]\to\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$. For $x\in[-1,1]$ and $\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, we write $\sin^{-1}x=\theta$ if $\sin\theta=x$; $\theta$ is called the principal value of $\sin^{-1}x$.

Ex.1 $\sin\dfrac{\pi}{6}=\dfrac12$, where $\dfrac12\in[-1,1]$ and $\dfrac{\pi}{6}\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, so $\sin^{-1}\dfrac12=\dfrac{\pi}{6}$: the principal value of $\sin^{-1}\dfrac12$ is $\dfrac{\pi}{6}$. Although $\sin\dfrac{5\pi}{6}=\dfrac12$ also, we cannot write $\sin^{-1}\dfrac12=\dfrac{5\pi}{6}$, since $\dfrac{5\pi}{6}\notin\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

Ex.2 $\sin\left(-\dfrac{\pi}{4}\right)=-\dfrac{1}{\sqrt2}$, where $-\dfrac{1}{\sqrt2}\in[-1,1]$ and $-\dfrac{\pi}{4}\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, so $\sin^{-1}\left(-\dfrac{1}{\sqrt2}\right)=-\dfrac{\pi}{4}$: the principal value of $\sin^{-1}\left(-\dfrac{1}{\sqrt2}\right)$ is $-\dfrac{\pi}{4}$. …