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:

\xy,(0,0);(40,0)**@{-};(60,30)**@{-};(0,0)**@{-},(20,-3)*{c},(7,2)*{\theta},(5% 0,12)*{a},(30,17)*{b}

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

\xy,(0,0);(40,0)**@{-};(60,30)**@{-};(0,0)**@{-},(20,-3)*{c},(7,2)*{\theta},(5% 0,12)*{a},(30,17)*{b},(40,0);(60,0)**@{--};(60,30)**@{--},(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

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

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

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

With this result in hand, we find that

$\displaystyle a^{2}$	$\displaystyle=$	$\displaystyle b^{2}-c^{2}-2cx$
$\displaystyle a^{2}$	$\displaystyle=$	$\displaystyle b^{2}-c^{2}-2c\left(b\cos\theta-c\right)$
$\displaystyle a^{2}$	$\displaystyle=$	$\displaystyle b^{2}-c^{2}-2bc\cos\theta+2c^{2}$
$\displaystyle a^{2}$	$\displaystyle=$	$\displaystyle b^{2}+c^{2}-2bc\cos\theta$	(1)