# What is cot(theta/2) in terms of trigonometric functions of a unit theta?

Feb 12, 2016

Sorry misread, $\cot \left(\setminus \frac{\theta}{2}\right) = \sin \frac{\setminus \theta}{1 - \cos \left(\setminus \theta\right)} ,$ which you can get from flipping $\tan \left(\setminus \frac{\theta}{2}\right) = \frac{1 - \cos \left(\setminus \theta\right)}{\sin} \left(\setminus \theta\right)$, proof coming.

$\setminus \theta = 2 \cdot \arctan \left(\frac{1}{x}\right)$

#### Explanation:

We cannot solve this without a right hand side, so I'm just going to go with $x$.

Goal rearranging, $\cot \left(\setminus \frac{\theta}{2}\right) = x$ for $\setminus \theta$.

Since most calculators or other aids don't have a "cot" button or a ${\cot}^{- 1}$ or $a r c \cot$ OR $a \cot$ button""^1 (different word for the inverse cotangent function, cot backward), we're going to do this in terms of tan.

$\cot \left(\setminus \frac{\theta}{2}\right) = \frac{1}{\tan} \left(\setminus \frac{\theta}{2}\right)$ leaving us with
$\frac{1}{\tan} \left(\setminus \frac{\theta}{2}\right) = x$ .

Now we take one over both sides.
$\frac{1}{\frac{1}{\tan} \left(\setminus \frac{\theta}{2}\right)} = \frac{1}{x}$ , which goes to

$\tan \left(\setminus \frac{\theta}{2}\right) = \frac{1}{x}$ .

At this point we need to get the $\setminus \theta$ outside of the $\tan$, we do this by taking the $\arctan ,$ the inverse of $\tan$. $\tan$ takes in an angle and produces a ratio, $\tan \left({45}^{o}\right) = 1$. $\arctan$ takes a ratio and produces an angle $\arctan \left(1\right) = {45}^{o}$ ""^2. This means that $\arctan \left(\tan \left(45\right)\right) = 45$ and $\tan \left(\arctan \left(1\right)\right) = 1$ or in general:

$\arctan \left(\tan \left(x\right)\right) = x$
and
$\tan \left(\arctan \left(x\right)\right) = x$.

Applying this to our expression we have,

$\arctan \left(\tan \left(\setminus \frac{\theta}{2}\right)\right) = \arctan \left(\frac{1}{x}\right)$ which becomes

$\setminus \frac{\theta}{2} = \arctan \left(\frac{1}{x}\right)$ and finishing up we get

$\setminus \theta = 2 \cdot \arctan \left(\frac{1}{x}\right)$.

You my notice I used footnotes! there are some subtleties to inverse trig functions I chose to pack down here.

1) Names of inverse trig functions. The formal name of an inverse trig function is "arc"- trig function ie. $\arctan$, $\arccos$ $\arcsin$. This is shorted two ways, "atan", "acos" "asin" which is used in computer programming and math programs and the HORRIBLE "tan^-1", "sin^-1" "cos^-1" which is used in a lot of calculators. It is HORRIBLE because ${\tan}^{-} 1 x$ can seem like $\frac{1}{\tan} x$, while $a \tan x$ and $\arctan x$ is much much less likely to confuse a reader. Use atan or arctan in your algebra.

2) Since all values of tangent occur TWICE in the unit circle, $\arctan$ normally returns angle between $- {180}^{o}$ and ${180}^{o}$, to use other angles you need to use your brain!