Arctangent of Root 3 over 3

Theorem

$\map \arctan {\dfrac {\sqrt 3} 3} = \dfrac \pi 6$

Proof

By definition, $\arctan$ is the inverse of the tangent function's restriction to $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.

By Tangent of $30 \degrees$:

$\tan \dfrac \pi 6 = \dfrac {\sqrt 3} 3$.

As $\dfrac \pi 6 \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$, we have by the definition of an inverse function:

$\map \arctan {\dfrac {\sqrt 3} 3} = \dfrac \pi 6$

$\blacksquare$