7. Given $\alpha, \beta \in [0, \pi]$. Then
$$
[\sin \alpha + \sin (\alpha + \beta)] \sin \beta
$$

the maximum value is . $\qquad$
7. $\frac{8 \sqrt{3}}{9}$.

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{[\sin \alpha+\sin (\alpha+\beta)]^{2}} \\
=(\sin \alpha+\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta)^{2} \\
=[\sin \alpha(1+\cos \beta)+\cos \alpha \cdot \sin \beta]^{2} \\
\leqslant\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)\left[(1+\cos \beta)^{2}+\sin ^{2} \beta\right] \\
=2+2 \cos \beta=4 \cos ^{2} \frac{\beta}{2} .
\end{array}
$$

Given $\beta \in[0, \pi]$, we have $\frac{\beta}{2} \in\left[0, \frac{\pi}{2}\right]$, thus,
$$
\begin{array}{l}
{[\sin \alpha+\sin (\alpha+\beta)] \sin \beta} \\
\leqslant 2 \sin \beta \cdot \cos \frac{\beta}{2}=4 \sin \frac{\beta}{2} \cdot \cos ^{2} \frac{\beta}{2} \\
=8 \sqrt{\sin ^{2} \frac{\beta}{2} \cdot \frac{1}{2} \cos ^{2} \frac{\beta}{2} \cdot \frac{1}{2} \cos ^{2} \frac{\beta}{2}} \\
\leqslant 8 \sqrt{\left(\frac{\sin ^{2} \frac{\beta}{2}+\frac{1}{2} \cos ^{2} \frac{\beta}{2}+\frac{1}{2} \cos ^{2} \frac{\beta}{2}}{3}\right)^{3}} \\
=\frac{8 \sqrt{3}}{9} .
\end{array}
$$

In the above two inequalities, equality holds if and only if
$$
\frac{\sin \alpha}{1+\cos \beta}=\frac{\cos \alpha}{\sin \beta} \text {, and } \sin ^{2} \frac{\beta}{2}=\frac{1}{2} \cos ^{2} \frac{\beta}{2}
$$

In this case, $\beta=2 \arctan \frac{\sqrt{2}}{2}, \alpha=\arctan \sqrt{2}$. Therefore, the maximum value sought is $\frac{8 \sqrt{3}}{9}$.