## Task B-3.1.

Let $x$ and $y$ be real numbers such that

$$
\sin x - \sin y = \sqrt{\frac{1008}{1009}}, \quad \cos x - \cos y = 1
$$

What is the value of $\cos (x-y)$?
## Solution.

After squaring the given equalities, we obtain:

$$
\begin{gathered}
\sin ^{2} x-2 \sin x \sin y+\sin ^{2} y=\frac{1008}{1009} \\
\cos ^{2} x-2 \cos x \cos y+\cos ^{2} y=1
\end{gathered}
$$

Let's add the obtained expressions.

$$
\begin{array}{cl}
\sin ^{2} x+\cos ^{2} x-2(\cos x \cos y+\sin x \sin y)+\sin ^{2} y+\cos ^{2} y=1+\frac{1008}{1009} & 1 \text { point } \\
1-2 \cos (x-y)+1=1+\frac{1008}{1009} & 2 \text { points }
\end{array}
$$

Thus, the final result is $\cos (x-y)=\frac{1}{2018}$.