How do you find the value of cot^-1 (-1/sqrt3)?

Aug 12, 2015

${\cot}^{- 1} \left(- \frac{1}{\sqrt{3}}\right) = {150}^{o} + n \cdot {180}^{o} = \frac{5 \pi}{6} + n \pi$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXXXXXXXXXXXXXXXXXX}}$for $\forall n \in \mathbb{Z}$

Explanation:

If $\theta = {\cot}^{- 1} \left(- \frac{1}{\sqrt{3}}\right)$
the
the ratio $\left(\text{opposite side")/("adjacent side}\right)$ of the defining triangle
must be
$\textcolor{w h i t e}{\text{XXXX}}$$\left(- 1\right) : \left(\sqrt{3}\right)$ (or equivalently $\left(1\right) : \left(- \sqrt{3}\right)$)

We therefore have the conditions below:

These are standard reference triangles and based on their quadrants the corresponding angles (within the range $\left[0 , {360}^{o}\right]$
are
$\textcolor{w h i t e}{\text{XXXX}}$${180}^{o} - {30}^{0} = {150}^{o} = \frac{5 \pi}{6}$ radians
and
$\textcolor{w h i t e}{\text{XXXX}}$${360}^{o} - {30}^{o} = {330}^{o} = \frac{11 \pi}{6}$ radians

These can be combined and include angles outside the $\left[0 , 2 \pi\right]$ range as
$\textcolor{w h i t e}{\text{XXXX}}$$\theta = {150}^{o} + n \cdot {180}^{o}$
or
$\textcolor{w h i t e}{\text{XXXX}}$$\theta = \frac{5 \pi}{6} + n \cdot \pi$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$for $\forall n \in \mathbb{Z}$