2. $\prod_{i=1}^{2018} \sin \frac{i \pi}{2019}=$ $\qquad$
2. $\frac{2019}{2^{2018}}$.

Since $z_{1}=\mathrm{e}^{\frac{2 \pi i}{n}}$ is a root of $z^{n}=1$, we have
$$
\begin{array}{l}
z^{n}-1=(z-1)\left(z-z_{1}\right)\left(z-z_{1}^{2}\right) \cdots\left(z-z_{1}^{n-1}\right) \\
\Rightarrow z^{n-1}+z^{n-2}+\cdots+z+1 \\
\quad=\left(z-z_{1}\right)\left(z-z_{1}^{2}\right) \cdots\left(z-z_{1}^{n-1}\right) .
\end{array}
$$

Let $z=1$, then
$$
\begin{array}{l}
n=\left(1-z_{1}\right)\left(1-z_{1}^{2}\right) \cdots\left(1-z_{1}^{n-1}\right) . \\
\text { Also, }\left|1-z_{1}^{k}\right|=\left|z_{1}^{-\frac{k}{2}}-z_{1}^{\frac{k}{2}}\right|=2 \sin \frac{k \pi}{n},
\end{array}
$$

Let $n=2019$, we get
$$
\prod_{i=1}^{2018} \sin \frac{i \pi}{2019}=\frac{2019}{2^{2018}} .
$$