I am failing to realize how do you go from
$\displaystyle cos(2t)+sin(2t)$
to
$\displaystyle \sqrt{2}cos(2t-\frac{\pi}{4})$
I was never good at trig.
$\displaystyle \displaystyle \sqrt{2}\cos{\left(2t - \frac{\pi}{4}\right)} = \sqrt{2}\left[\cos{(2t)}\cos{\left(\frac{\pi}{4}\right)} + \sin{(2t)}\sin{\left(\frac{\pi}{4}\right)}\right]$
$\displaystyle \displaystyle = \sqrt{2}\left[\frac{\cos{2t}}{\sqrt{2}} + \frac{\sin{2t}}{\sqrt{2}}\right]$
$\displaystyle \displaystyle = \cos{2t} + \sin{2t}$.