System Response
$ \cos(2t)\,= \frac{e^{2jt}+e^{-2jt}}{2}\, $.
We already had the response of $ e^{2jt}\, $ is $ te^{-2jt}\, $ and the response of $ e^{-2jt}\, $ is $ te^{2jt}\, $.
Assuming the system is a LTI system, we can substitute the response of $ \frac{e^{2jt}+e^{-2jt}}{2}\, $ with the values above.
Thus the system will produce the following response as an output :
$ \frac{te^{2jt}+te^{-2jt}}{2}\, $
$ =t\frac{e^{2jt}+e^{-2jt}}{2}\, $
$ =t\cos(2t)\, $
