Prove that:
(cos x+cos y)2+(sin x−sin y)2 = 4cos2(x+y2)
We have L.H.S
= (cos x+cos y)2+(sin x−sin y)2
= [2cos(x+y2)cos(x−y2)]2+[2cos(x+y2)sin(x−y2)]2
= 4cos2(x+y2)cos2(x−y2)+4cos2(x+y2)sin2(x−y2)
= 4cos2(x+y2)[cos2(x−y2)+sin2(x−y2)]
= 4cos2(x+y2) = R.H.S.