If cosxcosy= cosθ, prove that:

sin^2 (x/2) cos^2 (y/2) +cos^2 (x/2) sin^2 (y/2) = sin^2 (θ/2)

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- Mar 8th 2010, 08:13 AMyobaculTrigonometric Proof
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)

- Mar 8th 2010, 11:47 AMOpalg
- Mar 8th 2010, 09:07 PMyobaculNope ...
(Worried)

I tried to .... I managed to get what I required on the RHS but I didn't manage to rearrange the rest to get the LHS. - Mar 10th 2010, 04:58 AMOpalg
$\displaystyle \cos x\cos y = \cos\theta$

$\displaystyle (1-2\sin^2(x/2))(1-2\sin^2(y/2)) = 1-2\sin^2(\theta/2)$

$\displaystyle 1 - 2\sin^2(x/2) -2\sin^2(y/2) + 4\sin^2(x/2)\sin^2(y/2) = 1-2\sin^2(\theta/2)$

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.