The principal value of \(\sin^{−1}\left(\frac{−\sqrt{3}}{2}\right)\) is

Explanation:

If sinθ = x ⇒ θ = sin^-1x, for θ ∈ [-π/2, π/2]

sin (sin^-1 x) =x for -π/2 ≤ x ≤ π/2

We have, \(\sin^{−1}\left(\frac{−\sqrt{3}}{2}\right)\)

= sin^-1(sin( \(\frac{-\pi}{3}\)))  ----- Since \(\frac{-\pi}{3}\) ∈ [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)]

∴ sin-1(sin( \(\frac{-\sqrt{3}}{2}\))) = \(\frac{-\pi}{3}\)

Additional InformationPrincipal Values of Inverse Trigonometric Functions: