6. Given $\sin \alpha+\sin (\alpha+\beta)+\cos (\alpha+\beta)=\sqrt{3}, \beta \in\left[\frac{\pi}{4}, \pi\right]$, then $\beta=$
6. $\frac{\pi}{4}$ Detailed Explanation: Expanding $\sin \alpha+\sin (\alpha+\beta)+\cos (\alpha+\beta)=\sqrt{3}$, we get
$\sin \alpha(1+\cos \beta-\sin \beta)+\cos \alpha(\sin \beta+\cos \beta)=\sqrt{3}$. By the Cauchy-Schwarz inequality, we have
$3 \leqslant\left[(1+\cos \beta-\sin \beta)^{2}+(\sin \beta+\cos \beta)^{2}\right]\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right) \Rightarrow \cos \beta \geqslant \sin \beta$, and since
$\beta \in\left[\frac{\pi}{4}, \pi\right]$, then $\sin \beta \geqslant \cos \beta$, so $\beta=\frac{\pi}{4}$.