Where do I even start?
$\displaystyle 4(sin^2x)(cos^2x) + cos^2(2x) = 1$
please reread my post (note that i did not mention cos(2x)) and answer the question i asked
by the way, you squared your identity incorrectly: $\displaystyle (\cos^2 x - \sin^2 x)^2 {\color{red}\ne} \cos^4 x - \sin^4 x$, since we (should) know that $\displaystyle (a + b)^2 = a^2 + 2ab + b^2$
$\displaystyle \sin 2x = 2 \sin x \cos x$ so squared it becomes:
$\displaystyle 4\sin^2x\cos^2x$?
Oh I see...
So $\displaystyle \sin 2x = 2 \sin x \cos x + cos^2/2x = 1$
$\displaystyle \sin 2x = 2 \sin x \cos x + (cos2x)^2 = 1$
So $\displaystyle 2 \sin x \cos x + (cos^2 - sin^2)^2 = 1$