Question

Prove that $(cos\,x-cos\, y)^2 + (sin\,x – sin\,y)^2 = 4sin^2 \frac{x-y}{2}$

Answer 1

L.H.S. = (cos x+ cos y)2+(sin x-sin y)2

=cos2x+cos2y-2cos x cosy+sin2 x+sin2 y-2 sin x sin y

=(cos2x+sin2x)+(cos2y+sin2y)-2[cos x cos y+sin x sin y]

=1+1-2 [cos(x-y)] [cos(A-B)= cos A cosB+sin A sin B]

=2[1-cos(x-y)]

$=2[1-\\{1-2 sin^2(\frac{x-y}{2})\\}]\,\,\,[cos2A=1-2sin^2 A]$

$=4cos^2(\frac{x-y}{2})=R.H.S.$