I'm working with this:

$\displaystyle -\frac{3}{4}t $ $\displaystyle cos(2t) + \frac{3}{16}sin(4t)cos(2t) + cos(2t) + \frac{3}{8}sin^3(2t) + sin(2t)$

...and I could use some help understanding how this simplifies to some form of:

$\displaystyle Acos(2t) + Bsin(2t) + Ctcos(2t)$

I understand how the $\displaystyle sin(4t)$ breaks down into $\displaystyle 2sin(2t)cos(2t)$, but beyond that, I'm a bit lost.