4. If the functions $f(x)$ and $g(x)$ are defined on $\mathbf{R}$, and $f(x-y)=f(x) g(y)-g(x) f(y)$, if $f(-2)=f(1) \neq 0$, then $g(1)+g(-1)=$ (answer with a number).

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4. $\begin{array}{l} \because f(x-y)=f(x) g(y)-\dot{g}(x) f(y), \\ \therefore f(y-x)=f(y) g(x)-g(y) f(x)=-[f(x) g(y)-g(x) f(y)] . \\ \therefore f(x-y)=-f(y-x)=-f[-(x-y)], \\ \therefore f(-x)=-f(x), \text{ i.e., } f(x) \text{ is an odd function. } \\ \therefore f(1)=f(-2)=f(-1-1)=f(-1) g(1)-g(-1) f(1) \\ =-f(1)[g(1)+g(-1)], \\ \therefore g(1)+g(-1)=-1 .\end{array}$