What is the value of \(\sin^{-1}0\)?

Correct answer: \(0\)

Since \(\sin 0=0\) and \(0\) lies in the principal range of \(\sin^{-1}x\), which is \([-\frac{\pi}{2},\frac{\pi}{2}]\), we have \(\sin^{-1}0=0\). Note that although \(\sin\pi=0\), \(\pi\) is not in the principal range so arcsin does not return \(\pi\). Also \(\sin(\frac{\pi}{2})=1\) and \(\sin(-\frac{\pi}{2})=-1\), so those choices are incorrect. Exam tip: always recall the principal value range for \(\sin^{-1}x\) when evaluating inverse trig values.