Evaluate sin-1(sin 2)

Given sin^-1(sin 2)

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 = 2, which does not lie on the above range,

Therefore we know that sin (π – x) = sin (x)

Hence sin (π – 2) = sin (2) also π – 2 ∈ [\( (\dfrac{-\pi}{2},\dfrac{\pi}{2})\)]

Sin^-1(sin 2) = π – 2