Answer :

**To Prove:** 3 sin^{–1}x = sin^{–1} (3x – 4x^{3})**Proof: **

we know that,

sin3θ = 3 sinθ - 4 sin^{3}θ.....(1)

Let sinθ = x, then

θ = sin^{-1 }x

Put the value in equation 1, we get,

sin3θ = 3x - 4x^{3}

3θ = sin^{-1}(3x - 4x^{3})

3 sin^{–1}x = sin^{–1} (3x – 4x^{3})**Hence, Proved.**

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