2.2.11 * Given $\alpha, \beta \in\left[0, \frac{\pi}{4}\right]$. Then the maximum value of $\sin (\alpha-\beta)+2 \sin (\alpha+\beta)$ is . $\qquad$
Given that $\sin (\alpha-\beta)+2 \sin (\alpha+\beta)=3 \sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta$, by the Cauchy-Schwarz inequality, we have
$$
\text { the above expression } \leqslant \sqrt{(3 \sin \alpha)^{2}+\cos ^{2} \alpha} \cdot \sqrt{\cos ^{2} \beta+\sin ^{2} \beta}=\sqrt{8 \sin ^{2} \alpha+1} \leqslant \sqrt{5} \text {, }
$$

When $\alpha=\frac{\pi}{4}, \beta=\arcsin \frac{1}{\sqrt{10}}$, $\sin (\alpha-\beta)+2 \sin (\alpha+\beta)$ achieves its maximum value of $\sqrt{5}$.