Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 683: 111


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