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

$L.H.S. = (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2}$
$= (\cos^{2} x + \cos^{2} y + 2 \cos x \cos y) + (\sin^{2} x + \sin^{2} y - 2 \sin x \sin y)$
$= (\cos^{2} x + \sin^{2} x) + (\cos^{2} y + \sin^{2} y) + 2(\cos x \cos y - \sin x \sin y)$
$= 1 + 1 + 2 \cos(x + y) \quad [\text{Using } \cos(A + B) = \cos A \cos B - \sin A \sin B]$
$= 2 + 2 \cos(x + y)$
$= 2[1 + \cos(x + y)]$
$= 2 \left[1 + 2 \cos^{2} \left(\frac{x + y}{2}\right) - 1\right] \quad [\text{Using } \cos 2\theta = 2 \cos^{2} \theta - 1]$
$= 4 \cos^{2} \left(\frac{x + y}{2}\right) = R.H.S.$