Answer
$\frac{\sqrt{10}}{10}$
Work Step by Step
RECALL:
(1) $\theta = \arctan{x} \longrightarrow \tan{\theta}=x$
(2) Arctan is defined only ins $(-\frac{\pi}{2}, \frac{\pi}{2})$, which means the angle is either in Quadrant I or IV.
(3) $\tan{\theta} = \frac{y}{x}$
(4) $\cos{\theta} = \frac{x}{r}$, where $r=\sqrt{x^2+y^2}$
Let $\theta = \arctan{3}$
This means that $\tan{\theta} = 3$. Since tangent is positive, then the angle is in Quadrant I.
With $\tan{\theta} = 3=\frac{3}{1}=\frac{y}{x}$, we can set $y=3$ and $x=1$.
Solve for $r$ using the formula $r=\sqrt{x^2+y^2}$ to obtain:
$r=\sqrt{1^2+3^2}=\sqrt{1+9} =\sqrt{10}$
Thus,
$\cos{(\arctan{3})}
\\=\cos{\theta}
\\= \frac{x}{r}
\\= \frac{1}{\sqrt{10}}
\\= \frac{1}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}}
\\=\frac{\sqrt{10}}{10}$
