# What is Cos(arcsin(5/13) - arccos(7/25))?

$\setminus \frac{204}{325}$

#### Explanation:

$\setminus \cos \left(\setminus {\sin}^{- 1} \left(\frac{5}{13}\right) - \setminus {\cos}^{- 1} \left(\frac{7}{25}\right)\right)$

$= \setminus \cos \left(\setminus {\sin}^{- 1} \left(\frac{5}{13}\right)\right) \setminus \cos \left(\setminus {\cos}^{- 1} \left(\frac{7}{25}\right)\right) + \setminus \sin \left(\setminus {\sin}^{- 1} \left(\frac{5}{13}\right)\right) \setminus \sin \left(\setminus {\cos}^{- 1} \left(\frac{7}{25}\right)\right)$

$= \setminus \cos \left(\setminus {\cos}^{- 1} \left(\frac{12}{13}\right)\right) \left(\frac{7}{25}\right) + \left(\frac{5}{13}\right) \setminus \sin \left(\setminus {\sin}^{- 1} \left(\frac{24}{25}\right)\right)$

$= \left(\frac{12}{13}\right) \left(\frac{7}{25}\right) + \left(\frac{5}{13}\right) \left(\frac{24}{25}\right)$

$= \setminus \frac{84 + 120}{13 \setminus \cdot 25}$

$= \setminus \frac{204}{325}$

Jul 28, 2018

The answer is $= \frac{218}{325}$

#### Explanation:

Let ${\theta}_{1} = \arcsin \left(\frac{5}{13}\right)$

Then,

$\sin {\theta}_{1} = \frac{5}{13}$

And

$\cos {\theta}_{1} = \sqrt{1 - {\sin}^{2} {\theta}_{1}}$

$= \sqrt{1 - \frac{25}{169}}$

$= \sqrt{\frac{144}{169}}$

$= \frac{14}{13}$

Let ${\theta}_{2} = \arccos \left(\frac{7}{25}\right)$

$\cos {\theta}_{2} = \frac{7}{25}$

$\sin {\theta}_{2} = \sqrt{1 - {\cos}^{2} {\theta}_{2}}$

$= \sqrt{1 - \frac{49}{625}}$

$= \sqrt{\frac{576}{625}}$

$= \frac{24}{25}$

Therefore,

$\cos \left(\arcsin \left(\frac{5}{13}\right) - \arccos \left(\frac{7}{25}\right)\right)$

$= \cos \left({\theta}_{1} - {\theta}_{2}\right)$

$= \cos {\theta}_{1} \cos {\theta}_{2} + \sin {\theta}_{1} \sin {\theta}_{2}$

$= \frac{14}{13} \cdot \frac{7}{25} + \frac{5}{13} \cdot \frac{24}{25}$

$= \frac{218}{325}$