.
This question shows you are not familiar with the subject. My advice is that you should read your textbook/notes again. Work out few simple problems.And also, I'm confused about this.. is -tanx-tanx equal to -2tanx or -tan2x?
By the way -tan x -tan x = -2 tan x
I will answer your question, but I doubt you will understand every step clearly. First read the identities and know them clearly and then re read my post, you will understand the solution.
$\displaystyle \sin ^2 2x \cos^2 2x = \frac{\sin^2 4x}{4} = \frac14\left(\frac{1 - \cos 8x}{2}\right) = \frac18 (1 - \cos 8x)$
For the first equality, I have used the relation $\displaystyle \sin 2y = 2 \sin y \cos y$
For the second equality I have used $\displaystyle \cos 2y = 1 - 2\sin^2 y$