Prove that:
(cosx+cosy)2+(sinx−siny)2=4cos2x+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))
=4cos2x+y2=RHS...........(∵ 1+cos2x=2cos2x)
Hence proved