The principal value of \(\sin^{-1}\left(-\displaystyle\frac{1}{2}\right)\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\displaystyle\frac{\pi}{6}\)
The principal value branch of \(\sin^{-1}\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). We need the angle in this interval whose sine is \(-\frac{1}{2}\).
\( \displaystyle \sin\left(-\frac{\pi}{6}\right) = -\sin\frac{\pi}{6} = -\frac{1}{2} \)
and \(-\frac{\pi}{6}\) lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so
\( \displaystyle \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \)