## Task A-3.2.

Let $\alpha=\frac{2 \pi}{2021}$. Calculate $\cos \alpha \cdot \cos 2 \alpha \cdot \ldots \cdot \cos 1010 \alpha$
## Solution.

Let's denote the given expression by $P$.

In general, we know that $\sin 2 \varphi = 2 \sin \varphi \cos \varphi$, hence $\cos \varphi = \frac{\sin 2 \varphi}{2 \sin \varphi}$. Therefore, $P$ can be written in the following form:

$$
\begin{aligned}
P & = \cos \alpha \cdot \cos 2 \alpha \cdot \ldots \cdot \cos 1010 \alpha \\
& = \frac{\sin 2 \alpha}{2 \sin \alpha} \cdot \frac{\sin 4 \alpha}{2 \sin 2 \alpha} \cdot \ldots \cdot \frac{\sin 2020 \alpha}{2 \sin 1010 \alpha} \\
& = 2^{-1010} \cdot \frac{\sin 2 \alpha \cdot \sin 4 \alpha \cdot \ldots \cdot \sin 2020 \alpha}{\sin \alpha \cdot \sin 2 \alpha \cdot \ldots \cdot \sin 1010 \alpha}
\end{aligned}
$$

Notice that all expressions in the above equalities are well-defined: the term $\sin (k \alpha)$ is zero only when $k$ is a multiple of 2021.

By canceling the first 505 factors in the numerator with the corresponding factors in the denominator, we get

$$
\begin{aligned}
P & = 2^{-1010} \cdot \frac{\sin 1012 \alpha \cdot \sin 1014 \alpha \cdot \ldots \cdot \sin 2020 \alpha}{\sin \alpha \cdot \sin 3 \alpha \cdot \ldots \cdot \sin 1009 \alpha} \\
& = 2^{-1010} \cdot \frac{\sin 2020 \alpha \cdot \sin 2018 \alpha \cdot \ldots \cdot \sin 1012 \alpha}{\sin \alpha \cdot \sin 3 \alpha \cdot \ldots \cdot \sin 1009 \alpha}
\end{aligned}
$$

Since for $k=506,507, \ldots, 1010$ we have

$$
\sin (2 k \alpha) = \sin \left(\frac{4 k \pi}{2021}\right) = -\sin \left(2 \pi - \frac{4 k \pi}{2021}\right) = -\sin \left(2 \pi \frac{2021 - 2 k}{2021}\right) = -\sin ((2021 - 2 k) \alpha)
$$

we finally get

$$
\begin{aligned}
P & = 2^{-1010} \cdot \frac{\sin 2020 \alpha \cdot \sin 2018 \alpha \cdot \ldots \cdot \sin 1012 \alpha}{\sin \alpha \cdot \sin 3 \alpha \cdot \ldots \cdot \sin 1009 \alpha} \\
& = 2^{-1010} \cdot \frac{(-1) \cdot \sin \alpha \cdot (-1) \cdot \sin 3 \alpha \cdot \ldots \cdot (-1) \cdot \sin 1009 \alpha}{\sin \alpha \cdot \sin 3 \alpha \cdot \ldots \cdot \sin 1009 \alpha} \\
& = 2^{-1010} \cdot (-1)^{505} \\
& = -2^{1010}.
\end{aligned}
$$