Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.7 - The Dot Product - Exercise Set - Page 794: 87


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 $\sqrt{1-{{x}^{2}}}$.
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