9. Given that $\triangle A B C$ is an acute triangle, $b=2 c$, and $\sin B-\sin (A+B)=2 \sin C \cos A$. Then the range of $(\cos B+\sin B)^{2}+$ $\sin 2 C$ is $\qquad$ .
$$
\begin{array}{l}
b=2 c \Rightarrow \sin B=2 \sin C \Rightarrow 2 \sin C-\sin C=2 \sin C \cos A \Rightarrow A=60^{\circ} . \\
\text { Then } \sin \left(120^{\circ}-C\right)=2 \sin C \Rightarrow \frac{\sqrt{3}}{2} \cos C+\frac{1}{2} \sin C=2 \sin C \\
\Rightarrow \frac{\sqrt{3}}{2} \cos C=\frac{3}{2} \sin C \Rightarrow \tan C=\frac{\sqrt{3}}{3} \Rightarrow C=30^{\circ} \Rightarrow B=90^{\circ} .
\end{array}
$$

Without the condition that $\triangle A B C$ is an acute triangle, we have $(\cos B+\sin B)^{2}+\sin 2 C=1+\frac{\sqrt{3}}{2}$.