Answer
The algebraic expression of $\sin \left( {{\cos }^{-1}}x \right)$ is $\sqrt{1-{{x}^{2}}}$.
Work Step by Step
Let ${{\cos }^{-1}}x=\theta $. Then,
$\begin{align}
& \cos \theta =\frac{x}{1} \\
& =\frac{\text{perpendicular}}{\text{Hypotenuse}}
\end{align}$
According to the Pythagorean Theorem, to find out the perpendicular side,
$\text{Perpendicular}=\sqrt{{{\left( \text{Hypotenuse} \right)}^{2}}-{{\left( \text{Adjacent side} \right)}^{2}}}=\sqrt{{{\left( \text{1} \right)}^{2}}-{{\left( x \right)}^{2}}}=\sqrt{1-{{x}^{2}}}$
Now, consider
$\sin \left( {{\cos }^{-1}}x \right)=\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\sqrt{1-{{x}^{2}}}}{1}=\sqrt{1-{{x}^{2}}}$
Hence, the algebraic expression of $\sin \left( {{\cos }^{-1}}x \right)$ is $\sqrt{1-{{x}^{2}}}$.
