Question

What is the principal value of `sin^-1(-sqrt3/2)?`

Answer 1

Let `y=sin^-1(-sqrt3/2)`
Then,
\[\sin{y} = - \frac{\sqrt{3}}{2} = \sin\left( - \frac{\pi}{3} \right)\]
\[y = - \frac{\pi}{3} \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\]
Here
\[\left[ - \frac{\pi}{2}, \frac{\pi}{2} \right]\]  is the range of the principal value branch of inverse sine function.

∴ `sin^-1(-sqrt3/2)=-pi/3`