# How do you simplify sin ( sin^ -1 (-3/5) + tan^ -1(5/12)) ?

Oct 17, 2016

$\sin \left({\sin}^{- 1} \left(- \frac{3}{5}\right) + {\tan}^{- 1} \left(\frac{5}{12}\right)\right) = - \frac{16}{65}$

#### Explanation:

Let ${\sin}^{- 1} \left(- \frac{3}{5}\right) = \alpha$ and ${\tan}^{- 1} \left(\frac{5}{12}\right) = \beta$

then $\sin \alpha = - \frac{3}{5}$ and $\tan \beta = \frac{5}{12}$

and $\cos \alpha = \sqrt{1 - {\left(- \frac{3}{5}\right)}^{2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

$\cos \beta = \frac{1}{\sqrt{1 + {\tan}^{2} \beta}} = \frac{1}{\sqrt{1 + {\left(\frac{5}{12}\right)}^{2}}} = \frac{1}{\sqrt{1 + \frac{25}{144}}} = \frac{1}{\sqrt{\frac{169}{144}}} = \frac{1}{\frac{13}{12}} = \frac{12}{13}$

and $\sin \beta = \tan \beta \times \cos \beta = \frac{5}{12} \times \frac{1}{13} = \frac{5}{13}$

Hene, $\sin \left({\sin}^{- 1} \left(- \frac{3}{5}\right) + {\tan}^{- 1} \left(\frac{5}{12}\right)\right) = \sin \left(\alpha + \beta\right)$

= $\sin \alpha \cos \beta + \cos \alpha \sin \beta$

= $\left(- \frac{3}{5}\right) \times \frac{12}{13} + \frac{4}{5} \times \frac{5}{13}$

= $- \frac{36}{65} + \frac{20}{65} = - \frac{16}{65}$