## Precalculus (6th Edition) Blitzer

The algebraic expression of $\sin \left( {{\cos }^{-1}}x \right)$ is $\sqrt{1-{{x}^{2}}}$.
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 $\sqrt{1-{{x}^{2}}}$.