$\begin{align}  & {{\sin }^{-1}}x=2{{\cos }^{-1}}y \\ & \text{Let }\alpha ={{\sin }^{-1}}x,\text{ }x=\sin \alpha  \\ & \text{but }\cos \alpha =\sqrt{1-{{\sin }^{2}}\alpha }=\sqrt{1-{{x}^{2}}} \\ & \alpha ={{\cos }^{-1}}\sqrt{1-{{x}^{2}}} \\ & \tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{x}{\sqrt{1-{{x}^{2}}}} \\ & \alpha ={{\tan }^{-1}}\left( \frac{x}{\sqrt{1-{{x}^{2}}}} \right) \\ & \text{Also let }{{\cos }^{-1}}y=\beta  \\ & \text{        }y=\cos \beta  \\ & \text{But }\sin \beta =\sqrt{1-{{\cos }^{2}}\beta }=\sqrt{1-{{y}^{2}}} \\ & \tan \beta =\frac{\sin \beta }{\cos \beta } \\ & {{\sin }^{-1}}x=2{{\cos }^{-1}}y \\ & \text{Apply sine of a function to both sides} \\ & \sin ({{\sin }^{-1}}x)=\sin (2{{\cos }^{-1}}y) \\ & x=\sin 2\beta ({{\cos }^{-1}}y=\beta ) \\ & x=2\sin \beta \cos \beta  \\ & x=2\sin \left( {{\sin }^{-1}}\sqrt{1-{{y}^{2}}} \right)\cos \left( {{\sin }^{-1}}\sqrt{1-{{y}^{2}}} \right) \\ & x=2y\sqrt{1-{{y}^{2}}} \\ & \text{Square both sides} \\ & {{x}^{2}}=4{{y}^{2}}(1-{{y}^{2}}) \\\end{align}$