Given $3\sin^{-1}$ x = $\sin^{-1}$ (3x - $4x^{3}$) , x ∊ $\left[-\frac{1}{2},\frac{1}{2}\right]$ Let x = sin θ ⇒ θ = $\sin^{-1}$ x ... (i) ⇒ LHS = $\sin^{-1}$ (3 sin θ - 4 $\sin^{3}$ θ) = $\sin^{-1}$ (sin 3θ) (Since sin 3θ = 3 sin θ - 4 $\sin^{3}$ θ) = 3θ (Since $\sin^{-1}$ (sin x) = x) RHS = 3 $\sin^{-1}$ x = 3 $\sin^{-1}$ (sin θ) ... (ii) $3\sin^{-1}$ x = $\sin^{-1}(3x-4x^{3})$ From (i) and (ii)