I was reading out a book and it said to prove that \$y(t) = sin (t) x(t-2)\$ is time variant, so far of all the inputs I have tried, as well as the general input of giving a shift of T, the system seems to be time invariant.
This system is time variant because plugging in \$x(t-a)\$ does not equal \$y(t-a)\$.
For the first case you get:
However, if you offset the output by a you get:
Since \$y_1(t)\$ does not equal \$y_2(t)\$ the system is time variant. Look at the first example on this page if you are confused: https://en.wikipedia.org/wiki/Time-invariant_system
The system is time variant. Forget about the t-2 and think about what happens with sin(t) when you start the input at time t=0 as apposed to time t=pi/2 .
In both of these cases, let the input be a dirac delta function. In the first case, you should get nothing out, and in the second you would get the delta function back out.