# What is cos(arctan(-2) + arccos(5/13))?

Sep 24, 2015

$\cos \left(\arctan \left(- 2\right) + \arccos \left(\frac{5}{13}\right)\right) = \frac{29 \sqrt{5}}{65}$

#### Explanation:

First we expand that:
$\cos \left(\arctan \left(- 2\right) + \arccos \left(\frac{5}{13}\right)\right) =$
$\cos \left(\arctan \left(- 2\right)\right) \cos \left(\arccos \left(\frac{5}{13}\right)\right) - \sin \left(\arctan \left(- 2\right)\right) \sin \left(\arccos \left(\frac{5}{13}\right)\right)$

We know that $\cos \left(\arccos \left(\theta\right)\right) = \theta$ so we can rewrite that to
$\frac{5 \cos \left(\arctan \left(- 2\right)\right)}{13} - \sin \left(\arctan \left(- 2\right)\right) \sin \left(\arccos \left(\frac{5}{13}\right)\right)$

We know that ${\sin}^{2} \left(\theta\right) = 1 - {\cos}^{2} \left(\theta\right)$, so
${\sin}^{2} \left(\arccos \left(\frac{5}{13}\right)\right) = 1 - {\left(\frac{5}{13}\right)}^{2} = 1 - \frac{25}{169} = \frac{144}{169}$

Taking the root
$\sin \left(\arccos \left(\frac{5}{13}\right)\right) = \frac{12}{13}$
It's positive because the sine is always positive in the arccosine range. It can let us rewrite the first expansion to:
$\frac{5 \cos \left(\arctan \left(- 2\right)\right)}{13} - \frac{12 \sin \left(\arctan \left(- 2\right)\right)}{13}$

Now, from ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$, if we divide both sides by ${\cos}^{2} \left(\theta\right)$ we get ${\tan}^{2} \left(\theta\right) + 1 = \frac{1}{\cos} ^ 2 \left(\theta\right)$. So:
${\tan}^{2} \left(\arctan \left(- 2\right)\right) + 1 = \frac{1}{\cos} ^ 2 \left(\theta\right) \rightarrow {\cos}^{2} \left(\theta\right) = \frac{1}{{\left(- 2\right)}^{2} + 1}$
${\cos}^{2} \left(\theta\right) = \frac{1}{5} \rightarrow \cos \left(\theta\right) = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

It's positive because on the range of the arctangent, the cosine is always postive. So we can rewrite it to:
$\frac{\sqrt{5}}{13} - \frac{12 \sin \left(\arctan \left(- 2\right)\right)}{13}$

Since the tangent is negative, and the cosine positive, it means the sine is negative. Using ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$ we have that
${\sin}^{2} \left(\arctan \left(- 2\right)\right) = 1 - \frac{1}{5} = \frac{4}{5}$
$\sin \left(\arctan \left(- 2\right)\right) = - \frac{2}{\sqrt{5}} = - \frac{2 \sqrt{5}}{5}$

Which means we can rewrite it all to:
$\frac{\sqrt{5}}{13} - \frac{12 \left(- 2 \frac{\sqrt{5}}{5}\right)}{13}$

From there on it's algebra:
$\frac{5 \sqrt{5}}{65} + \frac{24 \sqrt{5}}{65} =$
$\frac{29 \sqrt{5}}{65}$

Sep 25, 2015

Find: $\cos \left(\arctan \left(- 2\right) + \arccos \left(\frac{5}{13}\right)\right)$

Ans: 1

#### Explanation:

Use calculator.
$\tan x = \left(- 2\right)$--> arc $x = - 63.43$ deg
$\cos y = \frac{5}{13} = 0.38$ --> arc $y = 67.38$ deg
cos (-63.43 + 67.58) = cos 3.95 = 0.997 = 1