11. Arrange the following 4 numbers in ascending order: $\sin \left(\cos \frac{3 \pi}{8}\right), \cos \left(\sin \frac{3}{8} \pi\right), \cos \left(\cos \frac{3}{8} \pi\right)$. $\sin \left(\sin \frac{3}{8} \pi\right)$ $\qquad$ .
11. Since when $x \in\left(0, \frac{\pi}{2}\right)$, $\cos x$ is monotonically decreasing, $\sin x$ is monotonically increasing, and $\sin x < \cos \frac{3}{8} \pi$, it follows that $\sin \left(\sin \frac{3}{8} \pi\right)>\sin \left(\cos \frac{3}{8} \pi\right), \cos \left(\sin \frac{3}{8} \pi\right) < \cos \frac{3}{8} \pi < \sin \left(\cos \frac{3}{8} \pi\right)$, and $\frac{3}{8} \pi > \frac{\pi}{4}$,
\[
\begin{aligned}
\text{so } \sin \left(\sin \frac{3 \pi}{8}\right) & > \cos \left(\sin \frac{3 \pi}{8}\right), \\
\cos \frac{3 \pi}{8} & = \sqrt{\frac{1+\cos \frac{3 \pi}{4}}{2}} \\
& = \frac{\sqrt{2-\sqrt{2}}}{2} < \sin \left(\sin \frac{3 \pi}{8}\right).
\end{aligned}
\]
Therefore, $\cos \left(\sin \frac{3 \pi}{8}\right) > \cos \frac{3}{8} \pi > \sin \left(\cos \frac{3 \pi}{8}\right)$.
Thus, $\cos \left(\cos \frac{3 \pi}{8}\right) > \sin \left(\sin \frac{3 \pi}{8}\right) > \cos \left(\sin \frac{3 \pi}{8}\right) > \sin \left(\cos \frac{3 \pi}{8}\right)$.