11. Given $\alpha, \beta \in\left[0, \frac{\pi}{4}\right]$. Then the maximum value of $\sin (\alpha-\beta)+$ $2 \sin (\alpha+\beta)$ is $\qquad$ .
11. $\sqrt{5}$.
$$
\begin{array}{l}
\text { Since } \sin (\alpha-\beta)+2 \sin (\alpha+\beta) \\
=3 \sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta,
\end{array}
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\text { the above expression } \leqslant \sqrt{(3 \sin \alpha)^{2}+\cos ^{2} \alpha} \cdot \sqrt{\cos ^{2} \beta+\sin ^{2} \beta} \\
=\sqrt{8 \sin ^{2} \alpha+1} \leqslant \sqrt{5} .
\end{array}
$$

When $\alpha=\frac{\pi}{4}, \beta=\arcsin \frac{1}{\sqrt{10}}$,
$$
\sin (\alpha-\beta)+2 \sin (\alpha+\beta)
$$

achieves the maximum value $\sqrt{5}$.