Given 3sin−1 x = sin−1 (3x - 4x3) , x ∊ [−

1

2

,

1

2

]
Let x = sin θ ⇒ θ = sin−1 x ... (i)
⇒ LHS = sin−1 (3 sin θ - 4 sin3 θ)
= sin−1 (sin 3θ) (Since sin 3θ = 3 sin θ - 4 sin3 θ)
= 3θ (Since sin−1 (sin x) = x)
RHS = 3 sin−1 x
= 3 sin−1 (sin θ) ... (ii)
3sin−1 x = sin−1(3x−4x3) From (i) and (ii)