Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 671: 104


The required value of the expression is $\frac{x\sqrt{1-{{y}^{2}}}-y}{\sqrt{{{x}^{2}}+1}}$.

Work Step by Step

Step 1: As, $\sin \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Hypotenuse}\left( \text{H} \right)}$ $\cos \theta =\frac{\text{Base}\left( \text{B} \right)}{\text{Hypotenuse}\left( \text{H} \right)}$ $\tan \theta =\frac{\text{Perpendicular}\left( \text{P} \right)}{\text{Base}\left( \text{B} \right)}$ Let us suppose ${{\tan }^{-1}}x=\alpha $ $\begin{align} & x=\tan \alpha \\ & \tan \alpha =\frac{x}{1} \\ & =\frac{\text{P}}{\text{B}} \end{align}$ By using the Pythagorian Theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$ $\begin{align} & {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\ & {{\text{H}}^{2}}={{x}^{2}}+{{1}^{2}} \\ & {{\text{H}}^{2}}={{x}^{2}}+1 \\ & \text{H}=\sqrt{{{x}^{2}}+1} \\ \end{align}$ Therefore, $\begin{align} & \sin \alpha =\frac{\text{P}}{\text{H}} \\ & =\frac{x}{\sqrt{{{x}^{2}}+1}} \end{align}$ Then, $\begin{align} & \cos \alpha =\frac{\text{B}}{\text{H}} \\ & =\frac{1}{\sqrt{{{x}^{2}}+1}} \end{align}$ Step 2: Similarly, ${{\sin }^{-1}}y=\beta $. This implies, $\begin{align} & y=\sin \beta \\ & \sin \beta =\frac{y}{1} \\ & =\frac{\text{P}}{\text{H}} \end{align}$ By using the Pythagorian Theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$ $\begin{align} & {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\ & {{1}^{2}}={{y}^{2}}+{{\text{B}}^{2}} \\ & {{\text{B}}^{2}}=1-{{y}^{2}} \\ & \text{B}=\sqrt{1-{{y}^{2}}} \\ \end{align}$ So, $\begin{align} & \cos \beta =\frac{\text{B}}{\text{H}} \\ & =\frac{\sqrt{1-{{y}^{2}}}}{1} \\ & =\sqrt{1-{{y}^{2}}} \end{align}$ Step 3: Using steps 1 and 2, $\sin \left( {{\tan }^{-1}}x-{{\sin }^{-1}}y \right)$ can be written as: $\sin \left( {{\tan }^{-1}}x-{{\sin }^{-1}}y \right)=\sin \left( \alpha -\beta \right)$ By using the trigonometric identity, $\sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta $ From above steps, $\sin \alpha =\frac{x}{\sqrt{{{x}^{2}}+1}}$ $\cos \alpha =\frac{1}{\sqrt{{{x}^{2}}+1}}$ $\cos \beta =\sqrt{1-{{y}^{2}}}$ $\sin \beta =y$ Thus, $\begin{align} & \sin \left( \alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ & =\left( \frac{x}{\sqrt{{{x}^{2}}+1}}\times \sqrt{1-{{y}^{2}}} \right)-\left( \frac{1}{\sqrt{{{x}^{2}}+1}}\times y \right) \\ & =\frac{x\sqrt{1-{{y}^{2}}}}{\sqrt{{{x}^{2}}+1}}-\frac{y}{\sqrt{{{x}^{2}}+1}} \\ & =\frac{x\sqrt{1-{{y}^{2}}}-y}{\sqrt{{{x}^{2}}+1}} \end{align}$
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