Question

Write the principal value of `sin^-1(-1/2)`

Answer 1

Let `y=sin^-1(-1/2)`

Then,

\[\sin{y} = - \frac{1}{2} = \sin\left( - \frac{\pi}{6} \right)\]
\[y = - \frac{\pi}{6} \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\]
Here, 
\[\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\]  is the range of the principal value branch of the inverse sine function.
∴ \[\sin^{- 1} \left( - \frac{1}{2} \right) = - \frac{\pi}{6}\]