I have to get this answer by tomorrow or I get an F, HELP!
here's the problem:

1-(2sinxcosx cos2x)^2
---------------------- <- over
2sin2xcos2x

(yes, it is a fraction)

the answer is supposed to be one and I have to prove it

I've gotten this far:

2-4sin^2xcox^2x 8sinxcos^6x-4sinxcosx 4cos^4x-4cosx
----------------------------------------------------
2sin2xcos2x

the rest is lost on me

The display is quite strange, I will interpret as
\frac{1-(2\sin x\cos x+\cos 2x)^2}{2\sin 2x\cos 2x}
(if you meant another expression, repost using "math" environment)

The problem is lot easier than what seems, just try to simplify as much as you can before going with pure computations. Otherwise you get an unmanageable expression.

Use first 2\sin x\cos x=\sin 2x

\frac{1-(2\sin x\cos x+\cos 2x)^2}{2\sin 2x\cos 2x}=\frac{1-(\sin 2x+\cos 2x)^2}{2\sin 2x\cos 2x}

Numerator:

1-(\sin 2x+\cos 2x)^2=1-(\sin^2 2x+\cos^2 2x +2\sin 2x\cos 2x)
and using \sin^2 2x+\cos^2 2x=1 ,
1-(\sin^2 2x+\cos^2 2x +2\sin 2x\cos 2x) =-2\sin 2x\cos 2x

Denominator is 2\sin 2x\cos 2x , thus the original expression

\frac{1-(2\sin x\cos x+\cos 2x)^2}{2\sin 2x\cos 2x}=-1