I have partially solved a question.
Prove that:
(sin x)^2 - (sin y)^2 = (sin x cos y)^2 - (cos x sin y)^2
$\displaystyle (\sin x)^2 - (\sin y)^2 = (\sin x \cos y)^2 - (\cos x \sin y)^2 $
$\displaystyle \sin^2 x - \sin^2 y = \sin^2 x \cos^2 y - \cos^2 x \sin^2 y$
$\displaystyle \sin^2 x - \sin^2 y = (1-cos^2 x)\cos^2 y - \cos^2 x (1-\cos^2 y)$
$\displaystyle \sin^2 x - \sin^2 y = \cos^2 y-cos^2 x\cos^2 y - \cos^2 x + \cos^2 x\cos^2 y$
$\displaystyle \sin^2 x - \sin^2 y = \cos^2 y - \cos^2 x$
$\displaystyle \sin^2 x - \sin^2 y = 1-\sin^2 y -(1- \sin^2 x)$
$\displaystyle \sin^2 x - \sin^2 y = 1-\sin^2 y -1+\sin^2 x$
$\displaystyle \sin^2 x - \sin^2 y = \sin^2 x-\sin^2 y$