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