Evaluate sin-1(sin 12)

Given sin^-1(sin 12)

We know that sin^-1(sin x) = x with x ∈ [\( (\dfrac{-\pi}{2},\dfrac{\pi}{2})\)] which is approximately equal to [-1.57, 1.57]

But here x = 12, which does not lie on the above range,

Therefore we know that sin (2nπ – x) = sin (-x)

Hence sin (2nπ – 12) = sin (-12)

Here n = 2 also 12 – 4π ∈ [\( (\dfrac{-\pi}{2},\dfrac{\pi}{2})\)]

Sin^-1(sin 12) = 12 – 4π