Question

Prove that: `(cosx-cosy)^2+(sinx-siny)^2=4sin^2``(x-y)/2`

Answer 1

LHS = `(cos x - cosy)^2 + (sin x - sin y)^2`
`= cos^2x + cos^2y - 2cosxcosy + sin^2x + sin^2y - 2sinxsiny`
`= 2 - 2[cosxcosy - sinx siny]`
using identity `cos(a+b) = cosacosb - sinasinb)`
`= 2-2cos(x-y)`
`= 2-2cos(2(x-y)/2)`
using identity `cos 2 theta = 1-2sin^2 theta`
`= 2-2[1-2sin^2((x-y)/2)]`
`= 2-2 +4sin^2((x-y)/2)`
`= 4sin^2((x-y)/2)`
=RHS
hence proved