Cosine rule

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 $c^2$.

$$2ab\cos(\gamma) = a^2 + b^2 - c^2$$

$$c^2 = a^2 + b^2 - 2ab\cos(\gamma)$$