# How do you evaluate tan^-1(-(sqrt(3)/3))?

We notice that for an angle $\theta$ we have that

$\tan \theta = - \frac{\sqrt{3}}{3} = - \frac{1}{\sqrt{3}} = - \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

Now we need to find the angles ${\theta}_{1}$,${\theta}_{2}$ which have

$\sin {\theta}_{1} = - \frac{1}{2}$ and $\cos {\theta}_{1} = \frac{\sqrt{3}}{2}$

and

$\sin {\theta}_{2} = \frac{1}{2}$ and $\cos {\theta}_{2} = - \frac{\sqrt{3}}{2}$

These angles are ${\theta}_{1} = {150}^{o} = 5 \cdot \frac{\pi}{6}$ and ${\theta}_{2} = {330}^{o} = 11 \cdot \frac{\pi}{6}$