2. If the functions $f(x)$ and $g(x)$ are defined on $\mathbf{R}$, and $f(x-y)=f(x) g(y)-$ $g(x) f(y), f(-2)=f(1) \neq 0$, then $g(1)+g(-1)=$ $\qquad$
2. -1 .
$f(y-x)=f(y) g(x)-g(y) f(x)=-f(x-y)=-f(-(y-x))$, so $f(x)$ is an odd function. Therefore, $f(1)=f(-2)=f(-1-1)=f(-1) g(1)-g(-1) f(1)=-f(1)[g(1)+g(-1)]$, thus $g(1)+g(-1)=-1$.