I have reasons to think that the dimension over $\mathbb{C}$ of $\mathbb{C} \{t\}$, the ring of (convergent) formal series modulo the ideal generated by $\sin(t)$, has finite dimension, how can I show this (or show that it is infinite dimensional).
The intuition comes from the identity $\sin(t)^2 + \cos(t)^2 = 1$, in some sense this "says" that $t = 1$.
The ring $A = \mathbb{C}[[t]]$ is local with maximal ideal $(t)$. Thus for any $f = a + tg$ with $a\in \mathbb{C}$ and $g\in A$, we have \begin{align*} f = a + \left(\frac{\sin t}{t}\right)^{-1}g \sin t. \end{align*} It follows that $A/(\sin t)$ is a $1$-dimensional vector space.