Answer
$2\sqrt{2}$ amperes.
Work Step by Step
We find critical points.
$i'(t)=-2\sin t+2\cos t$
$ i'(t)=0\quad$ for
$2\sin t=2\cos t$
$\tan t=1$
$t=\displaystyle \frac{\pi}{4}+k\pi,$ where k is a nonnegative integer.
Evaluating $i(t)$ at $t=\displaystyle \frac{\pi}{4},$
$i(\displaystyle \frac{\pi}{4})=2\cos(\frac{\pi}{4})+2\sin\frac{\pi}{4}=2(\frac{\sqrt{2}}{2})+2(\frac{\sqrt{2}}{2})=2\sqrt{2}$
Evaluating $i(t)$ at $t=\displaystyle \frac{5\pi}{4},$
$i(\displaystyle \frac{\pi}{4})=2\cos(\frac{5\pi}{4})+2\sin\frac{5\pi}{4}=2(-\frac{\sqrt{2}}{2})+2(-\frac{\sqrt{2}}{2})=-2\sqrt{2}$
The greatest magnitude (absolute value) is
$2\sqrt{2}$ amperes.