Law of Cosines/Proof 1

Theorem
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then $c^2 = a^2 + b^2 - 2ab \cos C$.

Proof
We can place this triangle onto a Cartesian coordinate system by plotting:


 * $A = \left({b \cos C, b \sin C}\right)$
 * $B = \left({a, 0}\right)$
 * $C = \left({0, 0}\right)$

By the distance formula, we have:
 * $c = \sqrt{\left({b \cos C - a}\right)^2 + \left({b \sin C - 0}\right)^2}$

Now, we just work with this equation: