Example 2 Given $m=\frac{\sin x}{\sin (y-z)}, n=\frac{\sin y}{\sin (z-x)}, p=\frac{\sin z}{\sin (x-y)}$, find the value of $m n + n p + p m$.
Solve $m n+n p+p m$
$$
=\frac{\sin x \sin y \sin (x-y)+\sin y \sin z \sin (y-z)+\sin z \sin x \sin (z-x)}{\sin (y-z) \sin (z-x) \sin (x-y)} \text {. }
$$

And $\sin x \sin y \sin (x-y)=\frac{1}{2}[\cos (x-y)-\cos (x+y)] \sin (x-y)$
$$
=\frac{1}{4}[\sin 2(x-y)-\sin 2 x+\sin 2 y] \text {. }
$$

Similarly, $\sin y \sin z \sin (y-z)=\frac{1}{4}[\sin 2(y-z)-\sin 2 y+\sin 2 z]$,
$$
\sin z \sin x \sin (z-x)=\frac{1}{4}[\sin 2(z-x)-\sin 2 z+\sin 2 x] .
$$

Adding the three equations yields $\frac{1}{4}[\sin 2(x-y)+\sin 2(y-z)+\sin 2(z-x)]=\frac{1}{2}[\sin (x-$
z) $\cos (x+z-2 y)+\sin (z-x) \cos (z-x)]=\frac{1}{2} \sin (z-x)[\cos (z-x)-\cos (x+z$
$-2 y)]=-\sin (y-z) \sin (z-x) \sin (x-y)$. Therefore, $m n+n p+p m=-1$.