If cos x cos y = cos θ, prove that:
sin^2 (x/2) cos^2 (y/2) +cos^2 (x/2) sin^2 (y/2) = sin^2 (θ/2)
$\displaystyle \cos x\cos y = \cos\theta$
$\displaystyle (1-2\sin^2(x/2))(1-2\sin^2(y/2)) = 1-2\sin^2(\theta/2)$
$\displaystyle 1 - 2\sin^2(x/2) -2\sin^2(y/2) + 4\sin^2(x/2)\sin^2(y/2) = 1-2\sin^2(\theta/2)$
Subtract 1 from both sides and divide by –2:
$\displaystyle \sin^2(x/2) +\sin^2(y/2) -2\sin^2(x/2)\sin^2(y/2) = \sin^2(\theta/2)$
Now write the left side of that as
$\displaystyle \sin^2(x/2)(1 - \sin^2(y/2)) + \sin^2(y/2)(1 - \sin^2(x/2))$
and you should be able to see how to complete the calculation from there.