# How do you evaluate arctan((sqrt (3)) /3)?

Apr 26, 2016

$\arctan \left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} = {30}^{\circ}$

#### Explanation:

Note that $\frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$
Here is a standard trigonometric triangle with this ratio for the tan:

Note that by definition the $\arctan$ function has a range of $\left[0 , \pi\right)$

Apr 26, 2016

$a r c \tan \left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6} \in \left[0 , \pi\right]$. It has two values $\frac{\pi}{6} \mathmr{and} \frac{7 \pi}{6} \in \left[0 , 2 \pi\right]$. The general value is $n \pi + \frac{\pi}{6} , n = 0 , \pm 1 , \pm 2 , \pm 3. . .$. ,

#### Explanation:

Thanks to Alan for the timely correction, over my seeing $\frac{\pi}{6} a s \frac{\pi}{3}$. Now, I am replacing $\frac{\pi}{3}$ by $\frac{\pi}{6}$, everywhere.
$\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3.}} \tan \left(\frac{7 \pi}{6}\right) = \tan \left(\pi + \frac{\pi}{6}\right) = \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

If a is a solution in $\left[0 , 2 \pi\right]$ for $b = \tan a$, then

$a r c \tan b = n \pi + a , n = 0 , \pm 1 , \pm 2 , \pm 3. . .$

Here, $b = \left(\frac{3}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}} \mathmr{and} a = \frac{\pi}{6} \mathmr{and} \frac{7 \pi}{6}$.