# 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)
Title derivation of cosines law DerivationOfCosinesLaw 2013-03-22 11:57:02 2013-03-22 11:57:02 drini (3) drini (3) 6 drini (3) Proof msc 51-00 CosinesLaw ProofOfCosinesLaw