# Question #0bee5

May 16, 2016

$\sin \left({\tan}^{-} 1 \left(\frac{3}{4}\right) + {\cos}^{-} 1 \left(\frac{5}{13}\right)\right) = \left(\frac{3}{5}\right) \left(\frac{5}{13}\right) + \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) = \frac{63}{65}$

#### Explanation:

${\tan}^{-} 1 \left(\frac{3}{4}\right)$ is a $\theta$ between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ with $\tan \theta = \frac{3}{4}$

${\cos}^{-} 1 \left(\frac{5}{13}\right)$ is a $\phi$ between $0$ and $\pi$ with $\cos \phi = \frac{5}{13}$

We have been asked to find $\sin \left(\theta + \phi\right)$.

We know that $\sin \left(\theta + \phi\right) = \sin \theta \cos \phi + \cos \theta \sin \phi$

We know that $\tan \theta = \frac{3}{4}$ and $0 < \theta < \frac{\pi}{2}$, we can find that $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$

We know that $\cos \phi = \frac{5}{13}$ and $0 < \phi < \frac{\pi}{2}$ so we find $\sin \phi = \frac{12}{13}$.

Therefore

$\sin \left({\tan}^{-} 1 \left(\frac{3}{4}\right) + {\cos}^{-} 1 \left(\frac{5}{13}\right)\right) = \left(\frac{3}{5}\right) \left(\frac{5}{13}\right) + \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) = \frac{63}{65}$ .