Prove that:
(cosx−cosy)2+(sinx−siny)2=4sin2x−y2
LHS=(cosx−cosy)2+(sinx−siny)2=cos2x+cos2y−2cosxcosy+sin2x+sin2y−2sinxsiny
=(cos2x+sin2x)+(cos2y+sin2y)−2(cosxcosy+sinxsiny)=1+1−2cos(x−y)
=2(1−cos(x−y))=4sin2x−y2=RHS.............(∵ 1−cos2x=2sin2x)
Hence proved