I am confused on the interpretation of $$ f(x)\delta(x-t_0) $$ where $\delta(x)$ is the Dirac delta function: $$ \delta(x-t_0)= \left\{ \begin{array}{ll} 0 & x< t_0 \\ \text{undefined} & x=t_0 \\ 0 & x>t_0\\ \end{array} \right. $$
The delta function is used in the discrete probability density functions, and more importantly in signal processing where the function is used as multiplier to represent discrete stuff.
Given,
$$ g(x)= \sin(x)\sum^n_{i=0}\delta(x-t_i) $$ What does $g(x)$ actually represent? If the delta function is undefined at $t_i$, what does it mean for it to be multiplied to $\sin(x)$?
I've seen that the graph of the given looks like a periodic graph of $\delta(x)$ at $t_i$ which seems to be enveloped by $\sin(x)$. How can this happen since the said function is undefined at all $t_i$?
The Dirac delta $\delta_{t_0}$ (where I write $\delta_{t_0}(x) = \delta(x-t_0)$) is defined not as you say but by its action on continuous test functions $\varphi\in C^0$. $$ \langle\delta_{t_0},\varphi\rangle = ∫_{\Bbb R} \varphi\, \delta_{t_0} = \varphi(t_0). $$ If you multiply it by a function $f$ then you have $$ \langle f\,\delta_{t_0},\varphi\rangle = ∫_{\Bbb R} \varphi\, f\,\delta_{t_0} = \varphi(t_0) \,f(t_0) = f(t_0) ∫_{\Bbb R} \varphi\,\delta_{t_0} = f(t_0)\,\langle \delta_{t_0},\varphi\rangle. $$ so you just have $f\,\delta_{t_0} = f(t_0)\,\delta_{t_0}$, i.e. you are just multiplying the delta by a constant. This is a Distribution similar to the delta but with integral $$ \int_{\Bbb R} f\,\delta_{t_0} = f(t_0). $$ (instead of $∫ \delta_{t_0} = 1$). Notice that the dirac Delta is not a function but a measure or a distribution of order $0$.
In your particular case, $$ f = \sin(x)\sum_i\delta_{t_i} = \sum_i \sin(x)\,\delta_{t_i} = \sum_i \sin(t_i)\,\delta_{t_i} $$ which indicates indeed that there are "masses" of size $\sin(t_i)$ at each $t_i$ and no mass outside of these points (which is sometimes represented as you say). I say masses because as you say, the Dirac delta is not a function so one can not talk about the value of the Dirac at one point. But around these points one has $\int_{t_i-\varepsilon}^{t_i+\varepsilon} f = \sin(t_i)$ if $\varepsilon$ is sufficenlty small