(i) Given as sin^-1(sin π/6)

As we know that the value of sin π/6 is ½

On, substituting these value in sin^-1(sin π/6)

We get, sin^-1 (1/2)

Let y = sin^-1 (1/2)

sin (π/6) = ½

So, the range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = ½

So, sin^-1(sin π/6) = π/6

(ii) Given as sin^-1(sin 7π/6)

We know that sin 7π/6 = – ½

On, substituting these in sin^-1(sin 7π/6) we get,

sin^-1 (-1/2)

Let y = sin^-1 (-1/2)

– sin y = ½

– sin (π/6) = ½

– sin (π/6) = sin (- π/6)

So, the range of principal value of sin^-1(-π/2, π/2) and sin (- π/6) = – ½

So, sin^-1(sin 7π/6) = – π/6

(iii) Given as sin^-1(sin 5π/6)

As we know that the value of sin 5π/6 is ½

Substitute this value in sin^-1(sin π/6)

We get, sin^-1 (1/2)

Let y = sin^-1 (1/2)

sin (π/6) = ½

So, the range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = ½

So, sin^-1(sin 5π/6) = π/6

(iv) Given as sin^-1(sin 13π/7)

The given trigonometry form can be written as sin (2π – π/7)

sin (2π – π/7) can be written as sin (π/7) [since sin (2π – θ) = sin (-θ)]

On, substituting these values in sin^-1(sin 13π/7) we get sin^-1(sin – π/7)

As sin^-1(sin x) = x with x ∈ [-π/2, π/2]

So, sin^-1(sin 13π/7) = – π/7

(v) Given as sin^-1(sin 17π/8)

Given can be written as sin (2π + π/8)

sin (2π + π/8) can be written as sin (π/8)

On, substituting these values in sin^-1(sin 17π/8) we get sin^-1(sin π/8)

As sin^-1(sin x) = x with x ∈ [-π/2, π/2]

So, sin^-1(sin 17π/8) = π/8