Proof of the cosine theorem

Hello everyone. Today we will prove the cosine theorem. We will use the method of direct proof as a proof method. This theorem was first formulated by the Persian mathematician Kashani. 

Theorem: Let $a, b, c$ be the sides of any triangle and let the angles $\alpha, \beta, \gamma$ be the angles opposite the sides $a, b, c$, respectively. Then the following equations hold: $$c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma$$ $$b ^ 2 = a ^ 2 + c ^ 2-2ac \cos \beta$$ $$a ^ 2 = b ^ 2 + c ^ 2-2bc \cos \alpha$$ Proof: 

We will now prove that the first equation holds i.e. : $$c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma$$ Other two equations can be proved in an analogous way. 

There are three possible cases, and these are: $\gamma$ is a right angle, $\gamma$ is an acute angle, and $\gamma$ is an obtuse angle. 

First case: $\gamma = 90 ^ {\circ}$ 

According to the Pythagoras' theorem, we know that for a right triangle with hypotenuse $c$ the following equation holds: $$c ^ 2 = a ^ 2 + b ^ 2$$ On the other hand, since $\cos \gamma = \cos 90 ^ {\circ} = 0$ it is true that: $$c ^ 2 = a ^ 2 + b ^ 2-0$$ $$c ^ 2 = a ^ 2 + b ^ 2-2ab \cos 90 ^ {\circ}$$ $$c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma$$ Second case: $\gamma <90 ^ {\circ}$ 

Consider the following diagram of an acute triangle $\triangle ABC$ with height $h$.

According to the Pythagoras' theorem, we can write the following two equations: $$a ^ 2 = h ^ 2 + p ^ 2$$ $$c ^ 2 = h ^ 2 + (bp) ^ 2$$ Combining these two equations we get: $$c ^ 2 = a ^ 2-p ^ 2 + (bp) ^ 2$$ $$c ^ 2 = a ^ 2-p ^ 2 + b ^ 2-2bp + p ^ 2$$ $$c ^ 2 = a ^ 2 + b ^ 2-2bp$$ Since $p = a \cos \gamma$ it follows: $$c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma$$Third case: $\gamma> 90 ^ {\circ}$ 

Consider the following diagram of an obtuse triangle $\triangle ABC$ with height $h$.

According to the Pythagoras' theorem, we can write the following two equations: $$a ^ 2 = h ^ 2 + p ^ 2$$ $$c ^ 2 = h ^ 2 + (b + p) ^ 2$$ Combining these two equations we get: $$c ^ 2 = a ^ 2-p ^ 2 + (b + p) ^ 2$$ $$c ^ 2 = a ^ 2-p ^ 2 + b ^ 2 + 2bp + p ^ 2$$ $$c ^ 2 = a ^ 2 + b ^ 2 + 2bp$$ Since $p = a \cos \left (180 ^{\circ}-\gamma \right)$ it follows: $$c ^ 2 = a ^ 2 + b ^ 2 + 2ab \cos \left (180 ^ {\circ}-\gamma \right)$$ Applying identity $\cos \left (180 ^ {\circ}-\gamma \right) = - \cos \gamma$ we get: $$c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma$$

$\blacksquare$