# What is Cos(arcsin(-5/13)+arccos(12/13))?

Jul 21, 2015

$= 1$

#### Explanation:

First you want to let $\alpha = \arcsin \left(- \frac{5}{13}\right)$ and $\beta = \arccos \left(\frac{12}{13}\right)$

So now we are looking for color(red)cos(alpha+beta)!

$\implies \sin \left(\alpha\right) = - \frac{5}{13} \text{ }$ and $\text{ } \cos \left(\beta\right) = \frac{12}{13}$

Recall : ${\cos}^{2} \left(\alpha\right) = 1 - {\sin}^{2} \left(\alpha\right) \implies \cos \left(\alpha\right) = \sqrt{1 - {\sin}^{2} \left(\alpha\right)}$

$\implies \cos \left(\alpha\right) = \sqrt{1 - {\left(- \frac{5}{13}\right)}^{2}} = \sqrt{\frac{169 - 25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$

Similarly, $\cos \left(\beta\right) = \frac{12}{13}$

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

$\implies \cos \left(\alpha + \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right)$

Then substitue all the values obtained ealier.

$\implies \cos \left(\alpha + \beta\right) = \frac{12}{13} \cdot \frac{12}{13} - \left(- \frac{5}{13}\right) \cdot \frac{5}{13} = \frac{144}{169} + \frac{25}{169} = \frac{169}{169} = \textcolor{b l u e}{1}$