# How do you simplify cos(arctan(x)) ?

Jul 26, 2016

$\frac{1}{\sqrt{1 + {x}^{2}}}$

#### Explanation:

Let simplify $\cos \left(A r c \tan \left(x\right)\right)$
Let $y = \arctan \left(x\right)$
$\iff x = \tan \left(y\right)$

$x = \sin \frac{y}{\cos} \left(y\right)$

We need to have an expression for $\cos \left(y\right)$ only,

${x}^{2} = \frac{\sin {\left(y\right)}^{2}}{\cos {\left(y\right)}^{2}}$

${x}^{2} + 1 = {\cancel{\sin {\left(y\right)}^{2} + \cos {\left(y\right)}^{2}}}^{= 1} / \cos {\left(y\right)}^{2}$

$\frac{1}{{x}^{2} + 1} = \cos {\left(y\right)}^{2}$

$\frac{1}{\sqrt{{x}^{2} + 1}} = \cos \left(y\right) = \cos \left(A r c \tan \left(x\right)\right)$

Aug 3, 2018

Continuation of my first answer that was edited by another.

#### Explanation:

Let a = arctan ( x ).

The principal value of a in ( − π/2, π/2 ).

Then $\tan a = x$ and

$0 \le \cos a \in \left[0 , 1\right)$ ( wrongly marked as [ 0, - 1 ], in my

Now, the given expression is

$\cos a = \frac{1}{\sqrt{1 + {x}^{2}}} , x \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$.

It is important that $\cos a \ge 0$, for $a \in {Q}_{1}$ or ${Q}_{4}$.

If the piecewise-wholesome general inverse operator

${\left(\tan\right)}^{- 1}$is used,

$\cos {\left(\tan\right)}^{- 1} x , = \pm \frac{1}{\sqrt{1 + {x}^{2}}}$

the negative sign is chosen, when $x \in {Q}_{3}$.

Example:

$\cos \left(\arctan 1\right) = \frac{1}{\sqrt{2}} , \arctan 1 = \frac{\pi}{4.}$

$\cos {\left(\tan\right)}^{- 1} 1 = \cos \left(k \pi + \frac{\pi}{4}\right) , k = 0 , \pm 1 , \pm 2 , \pm , \ldots$

$\in {Q}_{1}$ or ${Q}_{3} , x = \ldots \frac{\pi}{4} , \frac{5}{4} \pi , \ldots$

So, the value is $\pm \frac{1}{\sqrt{2}}$..

My intention, in this approach, is to inform about nuances..

So, I am making a second answer,