derivation of cosines law

The idea is to prove the cosines law:

$$a^2=b^2+c^2-2bc\cos \theta$$

where the variables are defined by the triangle:

$$,(0,0);(40,0)**\mathrm{@}-;(60,30)**\mathrm{@}-;(0,0)**\mathrm{@}-,(20,-3)*c,(7,2)*\theta ,(50,12)*a,(30,17)*b$$

Let’s add a couple of lines and two variables, to get

$$,(0,0);(40,0)**\mathrm{@}-;(60,30)**\mathrm{@}-;(0,0)**\mathrm{@}-,(20,-3)*c,(7,2)*\theta ,(50,12)*a,(30,17)*b,(40,0);(60,0)**\mathrm{@}--;(60,30)**\mathrm{@}--,(50,-3)*x,(63,15)*y$$

This is all we need. We can use Pythagoras’ theorem to show that

$$a^2=x^2+y^2$$

and

$$b^2=y^2+{\left(c+x\right)}^2$$

So, combining these two we get

$a^2$$=$$x^2+b^2-{\left(c+x\right)}^2$

$a^2$$=$$x^2+b^2-c^2-2cx-x^2$

$a^2$$=$$b^2-c^2-2cx$

So, all we need now is an expression for $x$. Well, we can use thedefinition of the cosine function to show that

$c+x$$=$$b\cos \theta$

$x$$=$$b\cos \theta -c$

With this result in hand, we find that

$a^2$$=$$b^2-c^2-2cx$

$a^2$$=$$b^2-c^2-2c\left(b\cos \theta -c\right)$

$a^2$$=$$b^2-c^2-2bc\cos \theta +2c^2$

$a^2$$=$$b^2+c^2-2bc\cos \theta$(1)