derivation of cosines law


The idea is to prove the cosines law:

a2=b2+c2-2bccosθ

where the variables are defined by the triangle:

{xy},(0,0);(40,0)**@-;(60,30)**@-;(0,0)**@-,(20,-3)*c,(7,2)*θ,(50,12)*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)*θ,(50,12)*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

a2=x2+y2

and

b2=y2+(c+x)2

So, combining these two we get

a2 = x2+b2-(c+x)2
a2 = x2+b2-c2-2cx-x2
a2 = b2-c2-2cx

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

c+x = bcosθ
x = bcosθ-c

With this result in hand, we find that

a2 = b2-c2-2cx
a2 = b2-c2-2c(bcosθ-c)
a2 = b2-c2-2bccosθ+2c2
a2 = b2+c2-2bccosθ (1)
Title derivation of cosines law
Canonical name DerivationOfCosinesLaw
Date of creation 2013-03-22 11:57:02
Last modified on 2013-03-22 11:57:02
Owner drini (3)
Last modified by drini (3)
Numerical id 6
Author drini (3)
Entry type Proof
Classification msc 51-00
Related topic CosinesLaw
Related topic ProofOfCosinesLaw