9. If the acute angle $\theta$ satisfies $\sin (\pi \cos \theta)=\cos (\pi \sin \theta)$, then $\sin 2 \theta=$
9. $\frac{3}{4}$.

Since $\theta \in\left(0, \frac{\pi}{2}\right)$, hence $0<\sin \theta<1$ and $0<\cos \theta<1$. Given that $\sin \theta>\frac{\sqrt{3}}{2}$, $\pi \cos \theta \in\left(\frac{\pi}{2}, \pi\right)$, $\sin (\pi \cos \theta)=\sin \left(\pi \cdot \cos \theta-\frac{\pi}{2}\right)$, and $\pi \cos \theta-\frac{\pi}{2} \in\left(0, \frac{\pi}{2}\right)$, so $\cos (\pi \sin \theta)=\sin (\pi \cos \theta)=\cos \left(\pi \cos \theta-\frac{\pi}{2}\right)$. Therefore, $\pi \sin \theta=\pi \cos \theta-\frac{\pi}{2}$, which means $\cos \theta-\sin \theta=\frac{1}{2}$. Squaring both sides gives $\sin 2 \theta=\frac{3}{4}$.