Couldn't think of a title, as i'll be using this thread more than once today for related, but different problems.

for starters:

$\displaystyle

\frac{1}{2}\sin{2x}[\tan^2{x}-\cos{2x}+2\cos^2{2x}]

$

For this, I ended up with -2sinxcosx, it's supposedly wrong.

Here's my work, for reference:

$\displaystyle

\frac{1}{2}\sin{2x}[\tan^2{x}-\cos{2x}+2\cos^2{2x}]

$

$\displaystyle

\frac{1}{2}\sin{2x}[\tan^2{x}-2\cos^2{x}-1+2\cos^2{2x}]

$

$\displaystyle

\frac{1}{2}\sin{2x}[\tan^2{x}-1]

$

$\displaystyle

\sin{x}\cos{x}(\tan^2{x}-1)

$

$\displaystyle

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

$

$\displaystyle

\frac{\sin{x}\cos{x}-\sin{x}\cos{x}\cos{2x}}{1+\cos{2x}}-\sin{x}\cos{x}

$

$\displaystyle

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

$

$\displaystyle

-2\sin{x}\cos{x}

$

Thanks in advance/