I'm trying to graph the function $f(t) = \sin(2(t-\pi))\mathcal{U}(t-\pi)$, where $\mathcal{U}$ is the Heaviside function defined by:
$\mathcal{U}(t-a) = \begin{cases} 0 &\mbox{} t \lt a \\ 1 & \mbox{} t \geq a \end{cases}$
I know that $\sin(2(t-\pi))=\sin(2t)$, so I figured $f$ is just a piecewise function that looks like
$f(t) = \begin{cases} 0 &\mbox{} t \lt \pi \\ \sin(2t)& \mbox{} t \geq \pi \end{cases}$
So then $f(t) = \sin(2(t-\pi))\mathcal{U}(t-\pi)$ would simply be the graph of $\sin(2t)$ after $t=\pi$ and $0$ everywhere else? I feel as though I am simplifying this too much.