Let $A, B, C$ be two interior angles of a triangle, then the minimum value of $\cos A(\sin B+\sin C)$ is $\qquad$
$\qquad$
8. $-\frac{2}{9} \sqrt{6}$.

If $A$ is a right angle, then $\cos A(\sin B-\sin C) \geqslant 0$; if $A$ is an obtuse angle, then $\cos A(\sin B-\sin C) \geqslant \frac{2}{9} \sqrt{6}$, equality holds when: $\cos ^{7} \frac{A}{2}=\frac{1}{6}$, i.e., $\cos A=-\frac{2}{3}, A=\pi-\operatorname{arcos} \frac{2}{3}, B=C$.