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

How do you simplify $\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)$?
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Melinda McCombs
Explanation:
Let simplify $\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)$
Let $y=\mathrm{arctan}\left(x\right)$
$x=\mathrm{tan}\left(y\right)$
$x=\frac{\mathrm{sin}\left(y\right)}{\mathrm{cos}\left(y\right)}$
We need to have an expression for $\mathrm{cos}\left(y\right)$ only,
${x}^{2}=\frac{{\mathrm{sin}\left(y\right)}^{2}}{{\mathrm{cos}\left(y\right)}^{2}}$
${x}^{2}+1=\frac{{\mathrm{sin}\left(y\right)}^{2}+{\mathrm{cos}\left(y\right)}^{2}}{{\mathrm{cos}\left(y\right)}^{2}}$
$\frac{1}{{x}^{2}+1}={\mathrm{cos}\left(y\right)}^{2}$
$\frac{1}{\sqrt{{x}^{2}+1}}=\mathrm{cos}\left(y\right)=\mathrm{cos}\left(\mathrm{arctan}\left(x\right)\right)$
###### Not exactly what you’re looking for?
Buck Henry
Explanation:
Let $a=\mathrm{arctan}\left(x\right)$
The principal value of $a\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
Then $\mathrm{tan}\alpha =x$ and
$0\le \mathrm{cos}\alpha \in \left[0,1\right)$ (wrongly marked as [0,-1], in my previous answer, two years ago).
Now the given expression is
$\mathrm{cos}\alpha =\frac{1}{\sqrt{1+{x}^{2}}},x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$
It is important that $\mathrm{cos}\alpha \ge 0$, for $\alpha \in {Q}_{1}$ or ${Q}_{4}$
If the piecewise-wholesome general inverse operator
$\left({\mathrm{tan}}^{-1}\right)$ is used
${\mathrm{cos}\left(\mathrm{tan}\right)}^{-1}x,=±\frac{1}{\sqrt{1+{x}^{2}}}$
the negative sign is chosen, when $x\in {Q}_{3}$
Example:
$\mathrm{cos}\left(\mathrm{arctan}1\right)=\frac{1}{\sqrt{2}},\mathrm{arctan}1=\frac{\pi }{4}$
${\mathrm{cos}\left(\mathrm{tan}\right)}^{-1}=\mathrm{cos}\left(k\pi +\frac{\pi }{4},k=0,±1,±2,±,\dots$
$\in {Q}_{1}$ or ${Q}_{3},x=\dots \frac{\pi }{4},\frac{5}{4}\pi ,\dots$
So, the value is $±\frac{1}{\sqrt{2}}$
My intention, in this approach, is to inform about nuances.