4A. Calculate the value of the expression

$$
\cos x \cos 2 x \cos 3 x \ldots \cos 1010 x \text {, for } x=\frac{2 \pi}{2021}
$$
Solution. Let us introduce the notations:

$$
P=\cos x \cos 2 x \cos 3 x \ldots \cos 1010 x \text { and } Q=\sin x \sin 2 x \sin 3 x \ldots \sin 1010 x
$$

We have

$2^{1010} P Q=2 \cos x \sin x \cdot 2 \cos 2 x \sin 2 x \cdot 2 \cos 3 x \sin 3 x \ldots 2 \cos 1010 x \sin 1010 x$

$=\sin 2 x \sin 4 x \sin 6 x \ldots \sin 1010 x \sin 1012 x \ldots \sin 2020 x$

$=\sin 2 x \sin 4 x \sin 6 x \ldots \sin 1010 x(-\sin (2 \pi-1012 x) \ldots(-\sin (2 \pi-2020 x))$

$=(-1)^{505} \sin 2 x \sin 4 x \sin 6 x \ldots \sin 1010 x \sin (2 \pi-1012 x) \ldots \sin (2 \pi-2020 x)$.

Now, if we use that for $x=\frac{2 \pi}{2021}$ we have

$$
\begin{aligned}
\sin (2 \pi-1012 x) & =\sin \left(2 \pi-1012 \frac{2 \pi}{2021}\right)=\sin 2 \pi\left(1-\frac{1012}{2021}\right) \\
& =\sin 1009 \cdot \frac{2 \pi}{2021}=\sin 1009 x
\end{aligned}
$$

and similarly

$$
\begin{aligned}
& \sin (2 \pi-1014 x)=\sin 1007 x \\
& \sin (2 \pi-1016 x)=\sin 1005 x \\
& \sin (2 \pi-2020 x)=\sin x
\end{aligned}
$$

we get

$$
\begin{aligned}
2^{1010} P Q & =(-1)^{505} \sin 2 x \sin 4 x \sin 6 x \ldots \sin 1010 x \sin 1009 x \sin 1007 x \ldots \sin x \\
& =-\sin x \sin 2 x \sin 3 x \ldots \sin 1010 x=-Q
\end{aligned}
$$

Furthermore, since $\sin k x=\sin \frac{2 k \pi}{2021} \neq 0, k=1,2, \ldots, 1010$, we get $Q \neq 0$, so from $2^{1010} P Q=-Q$ it follows that $P=-\frac{1}{2^{1010}}$.