(i) Given as sin^-1{(sin – 17π/8)}

As we know that – sin θ = sin (-θ)

So, (sin -17π/8) = – sin 17π/8

– sin 17π/8 = – sin (2π + π/8) [since sin (2π – θ) = sin (θ)]

It can also be written as – sin (π/8)

– sin (π/8) = sin (-π/8) [since – sin θ = sin (-θ)]

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 -π/8) = – π/8

(ii) Given as sin^-1(sin 3)

As we know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 3, which does not lie on the above range,

So, we know that sin (π – x) = sin (x)

Thus, sin (π – 3) = sin (3) also π – 3 ∈ [-π/2, π/2]

Sin^-1(sin 3) = π – 3

(iii) Given as sin^-1(sin 4)

As we know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 4, which does not lie on the above range,

So, we know that sin (π – x) = sin (x)

Hence sin (π – 4) = sin (4) also π – 4 ∈ [-π/2, π/2]

sin^-1(sin 4) = π – 4

(iv) Given as sin^-1(sin 12)

As we know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

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

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

Hence, sin (2nπ – 12) = sin (-12)

Here n = 2 also 12 – 4π ∈ [-π/2, π/2]

sin^-1(sin 12) = 12 – 4π

(v) Given as sin^-1(sin 2)

As we know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 2, which does not lie on the above range,

So, we know that sin (π – x) = sin (x)

Thus, sin (π – 2) = sin (2) also π – 2 ∈ [-π/2, π/2]

sin^-1(sin 2) = π – 2