11.1. It is known that

$$
\sin x \cos y=\cos x \sin y=\frac{1}{2}
$$

Find $\cos 2 x-\sin 2 y$.
Answer: -1 or 1.

First solution. Multiplying the given equations and multiplying the product by 4, we get $\sin 2 x \sin 2 y=1$. From this, $\sin 2 x=\sin 2 y=1$ or $\sin 2 x=\sin 2 y=-1$ (both cases are possible; it is sufficient to take $x=y=\frac{\pi}{4}$ or $x=y=\frac{3 \pi}{4}$). In both cases, $\cos 2 x=0$. Then $\cos 2 x-\sin 2 y=-1$ or $\cos 2 x-\sin 2 y=1$.

Second solution. From the condition, we have

$$
\sin (x+y)=\sin x \cos y+\sin y \cos x=1
$$

and, similarly, $\sin (x-y)=0$. Then $x+y=\frac{\pi}{2}+2 \pi k$ and $x-y=\pi \ell$ for integers $k$ and $\ell$. But then $2 x=(x+y)+(x-y)=\frac{\pi}{2}+\pi(2 k+\ell)$, from which $\cos 2 x=0$, and $2 y=(x+y)-(x-y)=\frac{\pi}{2}+\pi(2 k-\ell)$, from which $\sin 2 y= \pm 1$. Therefore, $\cos 2 x-\sin 2 y=\mp 1$. The same examples show that both answers are possible.

Comment. One of the answers is lost - deduct 2 points.

Examples of angles giving both answers are provided without justification - deduct 2 points.

An example of angles giving one answer is provided without justification - 0 points.