BITSAT Inverse Trigonometric Functions — practice questions

The principal-value range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. This is the unique interval on which $\sin$ is monotonic and covers its full range $[-1, 1]$. Quick reference: - $\sin^{-1}, \tan^{-1}, \csc^{-1}$: range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ (with closed/open variations). - $\cos^{-1}, \sec^{-1}, \cot^{-1}$: range $[0, \pi]$ (option A is for these).

$\sin\theta = \dfrac{1}{2}$ has the principal solution $\theta = \dfrac{\pi}{6}$ (30°), which lies in the standard range $[-\pi/2, \pi/2]$. Memorise the three classic values: $\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}$, $\sin^{-1}\dfrac{1}{\sqrt 2} = \dfrac{\pi}{4}$, $\sin^{-1}\dfrac{\sqrt 3}{2} = \dfrac{\pi}{3}$.

$\sin^{-1}$ is the inverse of $\sin$ restricted to its principal-value range $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. Inside that interval, $\sin^{-1}(\sin x)$ returns $x$ as expected. Outside that interval, $\sin^{-1}(\sin x)$ folds back to whatever principal value matches: e.g. $\sin^{-1}(\sin \pi) = \sin^{-1}(0) = 0$, not $\pi$. Identity $\sin(\sin^{-1} x) = x$ holds for all $x \in [-1, 1]$ (the domain side), no such restriction.