# How do you calculate sin(cos^-1(5/13)+tan^-1(3/4))?

May 18, 2018

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

#### Explanation:

Let ${\cos}^{- 1} \left(\frac{5}{13}\right) = x$ then

$\rightarrow \cos x = \frac{5}{13}$

$\rightarrow \sin x = \sqrt{1 - {\cos}^{2} x} = \sqrt{1 - {\left(\frac{5}{13}\right)}^{2}} = \frac{12}{13}$

$\rightarrow x = {\sin}^{- 1} \left(\frac{12}{13}\right) = {\cos}^{- 1} \left(\frac{5}{13}\right)$

Also, let ${\tan}^{- 1} \left(\frac{3}{4}\right) = y$ then

$\rightarrow \tan y = \frac{3}{4}$

$\rightarrow \sin y = \frac{1}{\csc} y = \frac{1}{\sqrt{1 + {\cot}^{2} y}} = \frac{1}{\sqrt{1 + {\left(\frac{4}{3}\right)}^{2}}} = \frac{3}{5}$

$\rightarrow y = {\tan}^{- 1} \left(\frac{3}{4}\right) = {\sin}^{- 1} \left(\frac{3}{5}\right)$

$\rightarrow {\cos}^{- 1} \left(\frac{5}{13}\right) + {\tan}^{- 1} \left(\frac{3}{4}\right)$

$= {\sin}^{- 1} \left(\frac{12}{13}\right) + {\sin}^{- 1} \left(\frac{3}{5}\right)$

$= {\sin}^{- 1} \left(\frac{12}{13} \cdot \sqrt{1 - {\left(\frac{3}{5}\right)}^{2}} + \frac{3}{5} \cdot \sqrt{1 - {\left(\frac{12}{13}\right)}^{2}}\right)$

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

Now, $\sin \left({\cos}^{- 1} \left(\frac{5}{13}\right) + {\tan}^{- 1} \left(\frac{3}{4}\right)\right)$

$= \sin \left({\sin}^{- 1} \left(\frac{63}{65}\right)\right) = \frac{63}{65}$