## Precalculus (6th Edition) Blitzer

The algebraic expression will be $2x\sqrt{1-{{x}^{2}}}$.
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 $2x\sqrt{1-{{x}^{2}}}$.