3. Find the minimum value of the expression

$$
A=\left(5\left(\cos x_{1}+\ldots+\cos x_{n}\right)+\sin x_{1}+\ldots+\sin x_{n}\right) \cdot\left(\cos x_{1}+\ldots+\cos x_{n}-5\left(\sin x_{1}+\ldots+\sin x_{n}\right)\right)
$$
Answer: $-13 n^{2}$.

Solution. For $k=1, \ldots, n$ let $t_{k}=3 \cos x_{k}-2 \sin x_{k}, s_{k}=2 \cos x_{k}+3 \sin x_{k}$. Then

$$
\begin{aligned}
A=\left(\sum_{k=1}^{n}\left(5 \cos x_{k}+\sin x_{k}\right)\right) \cdot\left(\sum_{k=1}^{n}\left(\cos x_{k}-5 \sin x_{k}\right)\right)= \\
=\left(\sum_{k=1}^{n}\left(3 \cos x_{k}-2 \sin x_{k}\right)+\sum_{k=1}^{n}\left(2 \cos x_{k}+3 \sin x_{k}\right)\right) \cdot\left(\sum_{k=1}^{n}\left(3 \cos x_{k}-2 \sin x_{k}\right)-\sum_{k=1}^{n}\left(2 \cos x_{k}+3 \sin x_{k}\right)\right)= \\
=\left(\sum_{k=1}^{n} t_{k}\right)^{2}-\left(\sum_{k=1}^{n} s_{k}\right)^{2}
\end{aligned}
$$

By the Cauchy inequality

$$
\left(\sum_{k=1}^{n} s_{k}\right)^{2} \leqslant n \sum_{k=1}^{n} s_{k}^{2} \leqslant n \sum_{k=1}^{n}\left(s_{k}^{2}+t_{k}^{2}\right)=n \sum_{k=1}^{n} 13\left(\cos ^{2} x_{k}+\sin ^{2} x_{k}\right)=13 n^{2}
$$

from which

$$
A \geqslant-\left(\sum_{k=1}^{n} s_{k}\right)^{2} \geqslant-13 n^{2}
$$

Equality is achieved when $\sin x_{1}=\ldots=\sin x_{n}=\frac{3}{\sqrt{13}}$ and $\cos x_{1}=\ldots=\cos x_{n}=\frac{2}{\sqrt{13}}$.