1. ## Need help simplifying

I'm working with this:

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

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

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

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

2. ## Re: Need help simplifying

Originally Posted by Lancet
I'm working with this:

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

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

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

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

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

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

$=-\frac{3t}{4}\cos(2t) + \frac{3}{8}\sin (2t)(1-\sin^{2}(2t)) + \cos(2t) + \frac{3}{8}\sin^{3}(2t) + \sin(2t)$

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

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

$=-\frac{3t}{4}\cos(2t)+\frac{11}{8}\sin (2t)+\cos(2t)$

Therefore, $A=1~,~B=\frac{11}{8}\mbox{ and }C=-\frac{3}{4}$

3. ## Re: Need help simplifying

Ah! Now I understand. Thank you for taking the time to explain it to me.