## Precalculus (6th Edition) Blitzer

The required value is $-\frac{33}{65}$
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 ${{\tan }^{-1}}\frac{4}{3}=\alpha$ \begin{align} & \frac{4}{3}=\tan \alpha \\ & \tan \alpha =\frac{4}{3} \\ & =\frac{\text{P}}{\text{B}} \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}} \\ & {{\text{H}}^{2}}={{4}^{2}}+{{3}^{2}} \\ & {{\text{H}}^{2}}=16+9 \\ & =25 \end{align} Now, taking the square root: \begin{align} & \text{H}=\sqrt{25} \\ & =5 \end{align} Then, \begin{align} & \,\cos \alpha =\frac{\text{B}}{\text{H}} \\ & =\frac{3}{5}\, \end{align} \begin{align} & \,\sin \alpha =\frac{\text{P}}{\text{H}} \\ & =\frac{4}{5} \end{align} Step 2: Similarly, ${{\cos }^{-1}}\frac{5}{13}=\beta$ this implies, \begin{align} & \frac{5}{13}=\cos \beta \\ & \cos \beta =\frac{5}{13} \\ & =\frac{\text{B}}{\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}} \\ & {{13}^{2}}={{\text{P}}^{2}}+{{5}^{2}} \\ & {{\text{P}}^{2}}=169-25 \\ & =144 \end{align} Now, taking the square root: \begin{align} & \text{P}=\sqrt{144} \\ & =12 \end{align} \begin{align} & \sin \beta =\frac{\text{P}}{\text{H}} \\ & =\frac{12}{13} \end{align} Step 3: Using steps 1 and 2, $\cos \left( ta{{n}^{-1}}\frac{4}{3}+{{\cos }^{-1}}\frac{5}{13} \right)$ can be written as: $\cos \left( ta{{n}^{-1}}\frac{4}{3}+{{\cos }^{-1}}\frac{5}{13} \right)=\cos \left( \alpha +\beta \right)$ By using the trigonometric identity, $\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$ From the above steps, $\sin \alpha =\frac{4}{5}$ $\cos \alpha =\frac{3}{5}$ $\cos \beta =\frac{5}{13}$ $\sin \beta =\frac{12}{13}$ \begin{align} & \,\cos \left( \alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & =\left( \frac{3}{5}\times \frac{5}{13} \right)-\left( \frac{4}{5}\times \frac{12}{13} \right) \\ & =\frac{15}{65}-\frac{48}{65} \\ & =-\frac{33}{65} \end{align}