Answer
The algebraic expression will be $2x\sqrt{1-{{x}^{2}}}$.
Work Step by Step
It is given that $x$ is positive and in the domain of the given inverse trigonometric function.
$x\in \left[ -1,1 \right]$ is also the domain of ${{\cos }^{-1}}x$.
So, in order to solve the equation, let:
$\begin{align}
& \alpha ={{\sin }^{-1}}x \\
& \sin \alpha =x
\end{align}$
$\sin \alpha $ is positive. That makes $\alpha $ be in the 1st quadrant where,
$\begin{align}
& \sin \alpha =x \\
& =\frac{x}{1}
\end{align}$
And the third side can be computed with the help of the Pythagorian Theorem:
$\begin{align}
& {{a}^{2}}+{{x}^{2}}={{1}^{2}} \\
& {{a}^{2}}=1-{{x}^{2}} \\
& a=\sqrt{1-{{x}^{2}}}
\end{align}$
And the value of $\cos \alpha $ will be:
$\begin{align}
& \cos \alpha =\frac{\sqrt{1-{{x}^{2}}}}{1} \\
& =\sqrt{1-{{x}^{2}}}
\end{align}$
And the value is computed as follows:
$\begin{align}
& \sin \left( 2{{\sin }^{-1}}x \right)=\sin 2y\text{ } \\
& =2\sin y\cdot \cos y \\
& =2\sin y\sqrt{1-{{x}^{2}}}
\end{align}$
By using $\left( {{\sin }^{-1}}y=x \right)$, the value becomes $2x\sqrt{1-{{x}^{2}}}$.
Thus, the algebraic expression will be $2x\sqrt{1-{{x}^{2}}}$.
