Evaluate sin-1{(sin – 17π/8)}

But we know that – sin θ = sin (-θ)

Therefore \( (sin -\dfrac{17\pi}{8}) = - sin( \dfrac{17\pi}{8})\)

\( - sin( \dfrac{17\pi}{8})= – sin (2π + \dfrac{\pi}{8})\) [since sin (2π – θ) = -sin (θ)]

It can also be written as \( – sin (\dfrac{\pi}{8})\)

\( – Sin (\dfrac{\pi}{8}) = sin (-\dfrac{\pi}{8})\) [since – sin θ = sin (-θ)]

By substituting these values in \( sin^{-1}\{ (sin -\dfrac{17\pi}{8}) \}\) we get,

\( sin^{-1}(sin – \dfrac{\pi}{8})\)

As sin^-1(sin x) = x with x ∈ [\( (\dfrac{-\pi}{2},\dfrac{\pi}{2})\)]

Therefore \( sin^{-1}(sin – \dfrac{\pi}{8})=\dfrac{-\pi}{8}\)