# Where does the cosine rule come from?

For the mathematical puritans out there the cosine rule derives from the following:

(you really don’t need to know this)

Find “a” KNOWING the angle A and length b and c.

Redraw the diagram

Pythagoras tells us that in triangle I , `x^2+d^2=b^2` ................ equation 1

Trigonometry tells us that `cosA=x/b`

Therefore `x=bcosA` ………… equation 2

Pythagoras also tells us that `(c-x)^2+d^2=a^2` …………… equation 3

Multiplying equation 3 out we get: `(c-x)(c-x)+d^2=a^2`

`c^2-cx-cx+x^2+d^2=a^2`

`c^2-2cx+x^2+d^2=a^2` ……………. equation 4

Substitute equation 1 into equation 4 we get:

`c^2-2cx+b^2=a^2`

Replacing `x` with equation 2 we get:

`c^2-2c(bcosA)+b^2=a^2`

Rearrange slightly

`a^2=b^2+c^2-2cb\cosA`

But again we emphasise that you really don’t need this.

# Where does the cosine rule come from?

For the mathematical puritans out there the cosine rule derives from the following:

(you really don’t need to know this)

Find “a” KNOWING the angle A and length b and c.

Redraw the diagram

Pythagoras tells us that in triangle Ã¢â€¦Â , `x^2+d^2=b^2` ................ equation 1

Trigonometry tells us that `cosA=x/b`

Therefore `x=bcosA` ………… equation 2

Pythagoras also tells us that `(c-x)^2+d^2=a^2` …………… equation 3

Multiplying equation 3 out we get: `(c-x)(c-x)+d^2=a^2`

`c^2-cx-cx+x^2+d^2=a^2`

`c^2-2cx+x^2+d^2=a^2` ……………. equation 4

Substitute equation 1 into equation 4 we get:

`c^2-2cx+b^2=a^2`

Replacing `x` with equation 2 we get:

`c^2-2c(bcosA)+b^2=a^2`

Rearrange slightly

`a^2=c^2+b^2-2c(bcosA)`

But again we emphasise that you really don’t need this.