# intrinsic cotangent formula

## Alec Jacobson

## January 23, 2012

Here's a derivation of the cotangent of an angle α of a triangle ABC given just the lengths of each side of the triangle.
First of all we remind ourselves what cotangent is in terms of cosine and sine:
```
cos α
cot α = ------
sin α
```

Now look at cosine and sine separately. By the law of cosines we have that
`a`^{2} = b^{2} + c^{2} - 2bc cos α

Rearranging things we have that
` -a`^{2} + b^{2} + c^{2}
cos α = ------------
2bc

Now looking at the sine, we start with the familiar area of the triangle treating b as the base.
```
1
A = --- bh
2
```

Using SOH-CAH-TOA, we replace the height:
```
1
A = --- bc sin α
2
```

Rearranging this we have:
```
2A
sin α = ----
bc
```

Now put the cosine and sine derivations together to get:
` cos α -a`^{2} + b^{2} + c^{2} bc
cot α = ------ = ------------ ----
sin α 2bc 2A

Finally arriving at:
` -a`^{2} + b^{2} + c^{2}
cot α = ------------
4A