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The principal value of  is

Answer (Detailed Solution Below)

Option 2 :

Detailed Solution

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Explanation:

If sinθ = x ⇒ θ = sin-1x, for θ ∈ [-π/2, π/2]

sin (sin-1 x) =x for -π/2 ≤ x ≤ π/2

We have, 

= sin-1(sin( ))  ----- Since  ∈ [, ]

∴ sin-1(sin( )) = 

Additional InformationPrincipal Values of Inverse Trigonometric Functions:

Function

Domain

Range of Principal Value

sin-1 x

[-1, 1]

[-π/2, π/2]

cos-1 x

[-1, 1]

[0, π]

csc-1 x

R - (-1, 1)

[-π/2, π/2] - {0}

sec-1 x

R - (-1, 1)

[0, π] - {π/2}

tan-1 x

R

(-π/2, π/2)

cot-1 x

R

(0, π)

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