9. In the tetrahedron $O A B C$, $\angle A O B=45^{\circ}, \angle A O C=\angle B O C=30^{\circ}$, then the cosine value of the dihedral angle $A-O C-B$'s plane angle $\alpha$ is $\qquad$.
As shown in the figure, construct $C D \perp$ plane $A O B, D A \perp O A$, $D B \perp O B$, then $O D$ bisects $\angle A O B$, $\triangle O A C \cong \triangle O B C$. Construct $A E \perp O C$, connect $B E$, it is easy to prove $B E \perp O C \Rightarrow \angle A E B$ is the plane angle of the dihedral angle $A-O C-B$. Let $O A=\sqrt{3} \Rightarrow A E=B E=\frac{\sqrt{3}}{2}, A B^{2}=3+3-2 \cdot 3 \cdot \frac{\sqrt{2}}{2}=6-3 \sqrt{2}$, in $\triangle A B E$, $\cos \angle A E B=\frac{\frac{3}{4}+\frac{3}{4}-6+3 \sqrt{2}}{2 \cdot \frac{3}{4}}=2 \sqrt{2}-3$.