`cos(A+B) = cosAcosB-sinAsinB`

Let `A=B=x/2`

`cos(A+B) = cosAcosB-sinAsinB`

`cos(x/2+x/2) = cos(x/2)cos(x/2)-sin(x/2)sin(x/2)`

`cosx = cos^2(x/2)-sin^2(x/2)`

But we know that `cos^2(x/2) = 1-sin^2(x/2)`

`cosx = cos^2(x/2)-sin^2(x/2)`

`cosx = 1-sin^2(x/2)-sin^2(x/2)`

`cosx = 1-2sin^2(x/2)`

We have `cos(A+B) = cosAcosB - sinAsinB`

Letting `y= A`

We have `cos(2y) = cos^2y - sin^2y = 1-sin^2y - sin^2y`

(using `cos^2A + sin^2A =1`)

This `implies`

`cos(2y) = 1-2sin^2y`

Letting `x = y/2` then this gives

`cosx = 1-2sin^2(x/2)` ** this completes the proof**