Given $\sin \alpha=\frac{5}{7}$, $\cos (\alpha+\beta)=\frac{11}{14}$, and $\alpha, \beta$ are acute angles, find $\cos \beta$.
Solving, we get $\cos \alpha=\frac{2 \sqrt{6}}{7}, \sin (\alpha+\beta)$ $=\frac{5 \sqrt{8}}{14}$, thus
$$
\begin{aligned}
\cos \beta & =\cos [(\alpha+\beta)-\alpha]=\cos (\alpha \\
& +\beta) \cos \alpha+\sin (\alpha+\beta) \sin \alpha \\
& =\frac{1}{98}(22 \sqrt{6}+25 \sqrt{3}) .
\end{aligned}
$$

Analysis. The result does not show any contradiction, but since $\frac{2 \sqrt{6}}{7}$
$$
\alpha+\beta, \beta<0$ contradicts the problem statement. It is evident that the original problem is incorrect. If we change it to $\cos (\alpha-\beta)=\frac{11}{14}$, the contradiction can be eliminated.