9.1. Find the maximum value of the expression $(\sqrt{8-4 \sqrt{3}} \sin x-3 \sqrt{2(1+\cos 2 x)}-2) \cdot(3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y)$. If the answer is not an integer, round it to the nearest integer.
Answer: 33. Solution. Let $f(x)=\sqrt{8-4 \sqrt{3}} \sin x-3 \sqrt{2(1+\cos 2 x)}-2$, $g(y)=3+2 \sqrt{11-\sqrt{3}} \cos y-\cos 2 y$ and estimate these two functions.

1) $f(x)=\sqrt{8-4 \sqrt{3}} \sin x-6|\cos x|-2 \leq \sqrt{8-4 \sqrt{3}} \cdot 1-6 \cdot 0-2=\sqrt{8-4 \sqrt{3}}-2=\sqrt{(\sqrt{6}-\sqrt{2})^{2}}-2$ $=\sqrt{6}-\sqrt{2}-2$. The specified value is achieved, for example, at $x=\frac{\pi}{2}$.

$f(x)=\sqrt{8-4 \sqrt{3}} \sin x-6|\cos x|-2=2(\sqrt{2-\sqrt{3}} \sin x-3|\cos x|-1) \geq 2\left(-\sqrt{(\sqrt{2-\sqrt{3}})^{2}+3^{2}}-1\right)$ $=-2(\sqrt{11-\sqrt{3}}+1)$. The specified value is achieved, for example, at $x=\arcsin \frac{3}{\sqrt{11-\sqrt{3}}}-\frac{\pi}{2}$.

Thus, $-2(\sqrt{11-\sqrt{3}}+1) \leq f(x) \leq \sqrt{6}-\sqrt{2}-2$. Note also that $-2(\sqrt{11-\sqrt{3}}+1)1$, so on the interval $t \in[-1 ; 1]$ the function $h(t)$ is increasing and
$h(-1) \leq h(t) \leq h(1)$. Further: $h(-1)=-2-2 \sqrt{11-\sqrt{3}}+4=-2(\sqrt{11-\sqrt{3}}-1)$,

$h(1)=-2+2 \sqrt{11-\sqrt{3}}+4=2(\sqrt{11-\sqrt{3}}+1)$. Here, $h(-1)<0<h(1)$.

Thus, $-2(\sqrt{11-\sqrt{3}}-1) \leq g(y) \leq 2(\sqrt{11-\sqrt{3}}+1)$.

3) Therefore, the maximum of the product is $(-2(\sqrt{11-\sqrt{3}}+1)) \cdot(-2(\sqrt{11-\sqrt{3}}-1))$ $=4(11-\sqrt{3}-1)=40-4 \sqrt{3} \approx 33$.