11. Given that the odd function $f(x)$ is a monotonically decreasing function on $[-1,0]$, and $\alpha, \beta$ are two interior angles of an acute triangle, the relationship in size between $f(\cos \alpha)$ and $f(\sin \beta)$ is $\qquad$ .
11. $f(\cos \alpha)>f(\sin \beta)$.

Given that $f(x)$ is a decreasing function on $[-1,0]$, it is also decreasing on $[0,1]$, and we have $\pi-(\alpha+\beta)\frac{\pi}{2}$. Therefore, $\frac{\pi}{2}>\beta>\frac{\pi}{2}-\alpha>0$, which implies $\sin \beta>\cos \alpha$, and $0<\sin \beta<1,0<\cos \alpha<1$. Hence, $f(\sin \beta)<f(\cos \alpha)$.