Evaluate sin-1(sin 7π/6)

But we know that  \( sin \dfrac{7\pi}{6} =-\dfrac{1}{2}\)

By substituting this in \( sin^{-1}( sin \dfrac{7\pi}{6})\) we get,

\(sin^{-1} ( \dfrac{-1}{2})\)

Now let y = \(sin^{-1} ( \dfrac{-1}{2})\)

– Sin y =\( \dfrac{1}{2}\)

 \(– Sin (\dfrac{\pi}{6}) = \dfrac{1}{2}\)

\( – Sin (\dfrac{\pi}{6}) = sin (\dfrac{-\pi}{6})\)

The range of principal value of \(sin^{-1} (\dfrac{-\pi}{2},\dfrac{\pi}{2})\) and  \(sin (\dfrac{-\pi}{6}) =- \dfrac{1}{2}\)

Therefore \( sin^{-1}( sin \dfrac{7\pi}{6})=\dfrac{-\pi}{6}\)