Question 52, in $\triangle ABC$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively, and $\sin A \sin 2A=$ $(1-\cos A)(1-\cos 2A)$, if the area of $\triangle ABC$ is $S=\frac{\sqrt{3}}{12}\left(8 b^{2}-9 a^{2}\right)$, find the value of $\cos B$.
The 52nd, Solution: According to the conditions, we have:
$$
\sin \mathrm{A} \sin 2 \mathrm{~A}=2 \sin ^{2} \mathrm{~A} \cos \mathrm{A}=(1-\cos \mathrm{A})(1-\cos 2 \mathrm{~A})=(1-\cos \mathrm{A}) \cdot 2 \sin ^{2} \mathrm{~A}
$$
$\Rightarrow \cos A=\frac{1}{2} \Rightarrow A=60^{\circ}$
Thus, we know:
$$
\begin{array}{l}
S=\frac{\sqrt{3}}{12}\left(8 \mathrm{~b}^{2}-9 \mathrm{a}^{2}\right)=\frac{b c}{2} \cdot \sin A=\frac{\sqrt{3 b c}}{4} \\
\Rightarrow 8 \mathrm{~b}^{2}-9 \mathrm{a}^{2}=3 \mathrm{bc}
\end{array}
$$

According to the cosine rule, we have:
$$
\begin{array}{l}
\mathrm{a}^{2}=\mathrm{b}^{2}+\mathrm{c}^{2}-2 \mathrm{bc} \cos \mathrm{A}=\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{bc} \\
\text { From (1) and (2), we know: }
\end{array}
$$

From (1) and (2), we know:
$$
\begin{array}{l}
8 b^{2}-9\left(b^{2}+c^{2}-b c\right)=3 b c \\
\Rightarrow(b-3 c)^{2}=0 \\
\Rightarrow b=3 c
\end{array}
$$

From (2) and (3), we know:
$$
\begin{array}{l}
a^{2}=b^{2}+c^{2}-b c=7 c^{2} \\
\Rightarrow a=\sqrt{7} c
\end{array}
$$

From (3), (4), and the cosine rule, we know:
$$
\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 b c}=\frac{(\sqrt{7} c)^{2}+c^{2}-(3 c)^{2}}{2 \cdot \sqrt{7} \cdot c}=-\frac{\sqrt{7}}{14}
$$

In conclusion, the value of $\cos B$ is $-\frac{\sqrt{7}}{14}$.