Statement
In a triangle, the following equation applies for every angle:
$$ c^2 = a^2 + b^2 - 2ab\cos(\gamma) $$
Proof
Construct altitude $ h $ and divide side $ a $ in $ a_1 $ and $ a_2 $, like the image below.
From the definition of cosine follows that $ \cos(\gamma) = \frac{a_1}{b} $.
Note that $ a = a_1 + a_2 $, which implies that $ a_2 = a - a_1 $.
Using the Pythagorean theorem follow these equations:
$$ b^2 = a_1^2 + h^2 \implies h^2 = b^2 - a_1^2 $$
$$ c^2 = a_2^2 + h^2 \implies h^2 = c^2 - a_2^2 $$
Now combine these equations.
$$ b^2 - a_1^2 = c^2 - a_2^2 $$
Substitute $ a_2 = a - a_1 $ and expand the brackets.
$$ b^2 - a_1^2 = c^2 - (a - a_1)^2 $$
$$ b^2 - a_1^2 = c^2 - (a^2 - 2 * a * a_1 + a_1^2) $$
$$ b^2 - a_1^2 = c^2 - a^2 + 2 * a * a_1 - a_1^2 $$
Cancel like terms and isolate $ a_1 $.
$$ b^2 = c^2 - a^2 + 2 * a * a_1 $$
$$ 2 * a * a_1 = a^2 + b^2 - c^2 $$
$$ a_1 = \frac{a^2 + b^2 - c^2}{2a} $$
Now substitue $ a_1 $ in $ \cos(\gamma) = \frac{a_1}{b} $ and remove the complex fraction.
$$ \cos(\gamma) = \frac{\frac{a^2 + b^2 - c^2}{2a}}{b} = \frac{a^2 + b^2 - c^2}{2ab} $$
Finally, isolate $ c^2 $.
$$ 2ab\cos(\gamma) = a^2 + b^2 - c^2 $$
$$ c^2 = a^2 + b^2 - 2ab\cos(\gamma) $$