## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 671: 102

#### Answer

$-1$

#### 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 ${{\cos }^{-1}}\left( -\frac{\sqrt{3}}{2} \right)=\alpha$ \begin{align} & \frac{-\sqrt{3}}{2}=\cos \alpha \\ & \cos \alpha =\frac{-\sqrt{3}}{2} \\ & =\frac{B}{H} \end{align} By using the Pythagoras theorem, ${{\left( \text{Hypotenuse} \right)}^{2}}=\text{ }{{\left( \text{Perpendicular} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}$ \begin{align} & {{\text{H}}^{2}}={{\text{P}}^{2}}+{{\text{B}}^{2}} \\ & {{P}^{2}}={{H}^{2}}-{{B}^{2}} \\ & {{P}^{2}}=4-3 \\ & =1 \end{align} Now, taking the square root: \begin{align} & P=\sqrt{1} \\ & =1 \end{align} Then, \begin{align} & \,\tan \alpha =\frac{P}{B} \\ & =-\frac{1}{\sqrt{3}}\, \end{align} \begin{align} & \,\sin \alpha =\frac{\text{P}}{\text{H}} \\ & =\frac{1}{2} \end{align} Step 2: Similarly, ${{\sin }^{-1}}\left( -\frac{1}{2} \right)=\beta$ this implies, \begin{align} & -\frac{1}{2}=\sin \beta \\ & \sin \beta =-\frac{1}{2} \\ & =\frac{P}{\text{H}} \end{align} By using the Pythagoras 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{B}}^{2}}={{\text{H}}^{2}}\text{-}{{\text{P}}^{2}} \\ & {{B}^{2}}={{4}^{2}}-{{1}^{2}} \\ & {{B}^{2}}=3 \end{align} Now, taking the square root: $B=\sqrt{3}$ \begin{align} & \cos \beta =\frac{B}{H} \\ & =\frac{\sqrt{3}}{2} \end{align} Step 3: Using steps 1 and 2, $\cos \left[ {{\cos }^{-1}}\left( -\frac{\sqrt{3}}{2} \right)-{{\sin }^{-1}}\left( -\frac{1}{2} \right) \right]$ can be written as: $\cos \left[ {{\cos }^{-1}}\left( -\frac{\sqrt{3}}{2} \right)-{{\sin }^{-1}}\left( -\frac{1}{2} \right) \right]=\cos \left( \alpha -\beta \right)$ By using the trigonometric identity, $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$ From the above steps, $\sin \alpha =\frac{1}{2}$ $\cos \alpha =-\frac{\sqrt{3}}{2}$ $\cos \beta =\frac{\sqrt{3}}{2}$ $\sin \beta =-\frac{1}{2}$ \begin{align} & \,\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ & =\left( -\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2} \right)+\left( \frac{1}{2}\times -\frac{1}{2} \right) \\ & =-\frac{3}{4}-\frac{1}{4} \\ & =\frac{-4}{4}=-1 \end{align}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.