An interesting Legendre symbol identity: $\left (\frac{a}{p} \right ) = \prod_{h\in\mathscr{H}}^{}\frac{\sin(2\pi ah/p)}{\sin(2\pi h/p)}$

Question

If we call $\mathscr{H}$ a one-half set of reduced residues (mod $p$), $p$ is a prime, if $\mathscr{H}$ has the property that:

$h \in \mathscr{H}$ if and only if $-h \notin \mathscr{H}$

Let $\mathscr{H}$ and $\mathscr{K}$ be two complementary one-half sets. They form the reduced residue system modulo $p$.

There are some propositions about $\mathscr{H}$ and $\mathscr{K}$:

1. if $(a, p) = 1$. Let $\nu$ be the number of $h \in \mathscr{H}$ for which $ah \notin \mathscr{H}$. That is $h \in \mathscr{H}$ but $ah \in \mathscr{K}$.

   Then we can get: \begin{split} (-1)^{\nu} = \left(\frac{a}{p}\right) \end{split} $\left(\frac{a}{p}\right)$ is the Legendre symbol

2. $(a, p) = 1$, $a \mathscr{H}$ and $a\mathscr{K}$ are complementary one-half sets.

   That is $a \mathscr{H}$ and $a\mathscr{K}$ are disjoint and form the reduced residue system modulo $p$.

I wonder how to get the following equation:

\begin{split} \left(\frac{a}{p}\right) = \prod_{h\in\mathscr{H}}^{}\frac{\sin(2\pi ah/p)}{\sin(2\pi h/p)} \end{split} for any integer $a$ and odd prime $p$.

Here are the hints:

Notice that $a\mathscr{H} = (a\mathscr{H} \cap \mathscr{H}) \cup(a\mathscr{H} \cap \mathscr{K})$.

Consider the product of elements in $\mathscr{H}$ and $(\mathscr{H} \cap a\mathscr{H}) \cup(\mathscr{H} \cap a\mathscr{K})$ respectively, and they are equal.