Example 2 Let the quadratic function $y=-x^{2}+(m-2) x +3(m+1)$ have its graph intersecting the $x$-axis at points $A$ and $B$ ($A$ to the left of $B$), and the $y$-axis at point $C$. The product of the lengths of segments $AO$ and $OB$ equals 6 ($O$ is the origin). Connect $AC$ and $BC$. Find the value of $\sin C$.
(Harbin 16th Junior High School Mathematics Competition)
Solution: According to the problem, we have $A O \cdot O B=6$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$, then $A O=\left|x_{1}\right|, O B=\left|x_{2}\right|$. Therefore, $\left|x_{1}\right| \cdot\left|x_{2}\right|=6$, which means $\left|x_{1} x_{2}\right|=6$.
We consider two cases:
(1) $x_{1} x_{2}=6$,

i.e., $x_{1} x_{2}=-3\left(n_{i}+1\right)=6$,
we have $m=-3, y=-x^{2}-5 x-6$.
Let $y=0$, solving gives $x_{1}=-3, x_{2}=-2$.
Thus, $A(-3,0), B(-2,0), C(0,-6)$.

Therefore,
$$
\begin{aligned}
S_{\triangle A B C} & =\frac{1}{2} A B \cdot O C \\
& =\frac{1}{2}(-2+3) \times|-6|=3, \\
A C & =\sqrt{9+36}=3 \sqrt{5}, \\
B C & =\sqrt{4+36}=2 \sqrt{10} .
\end{aligned}
$$

Hence $\frac{1}{2} \cdot 3 \sqrt{5} \cdot 2 \sqrt{10} \cdot \sin C=3$,
$$
\sin C=\frac{1}{5 \sqrt{2}}=\frac{\sqrt{2}}{10} \text {. }
$$
(2) $x_{1} x_{2}=-6$,

i.e.,
$$
\begin{aligned}
x_{1} x_{2} & =-3(m+1) \\
& =-6,
\end{aligned}
$$

we have $m=1, y=-x^{2}-x+6$.
Let $y=0$, solving gives $x_{1}=-3, x_{2}=2$.
Thus, $A(-3,0), B(2,0), C(0,6)$.
Therefore, $S_{\triangle A B C}=\frac{1}{2}(2+3) \times 6=15$,
$$
\begin{array}{l}
A C=\sqrt{9+36}=3 \sqrt{5}, \\
B C=\sqrt{4+36}=2 \sqrt{10} . \\
\text { Hence } \frac{1}{2} \cdot 3 \sqrt{5} \cdot 2 \sqrt{10} \cdot \sin C=15, \\
\sin C=\frac{\sqrt{2}}{2} .
\end{array}
$$

In summary, $\sin C=\frac{\sqrt{2}}{10}$ or $\sin C=\frac{\sqrt{2}}{2}$.