3.1. Compute the product

$$
\cos a \cdot \cos 2a \cdot \cos 3a \cdot \ldots \cdot \cos 999a
$$

where $a=2\pi / 1999$.
3.1. Answer: $2^{-999}$.

Let the desired product be denoted by $P$. Using the formula $\sin 2 x=2 \sin x \cos x$, we get

$$
\sin 2 a \cdot \sin 4 a \cdot \sin 6 a \cdot \ldots \cdot \sin 1998 a=2^{999} \cdot P \cdot \sin a \cdot \sin 2 a \cdot \ldots \cdot \sin 999 a
$$

On the other hand, $\sin 1998 a=-\sin a, \sin 1996 a=-\sin 3 a, \ldots, \sin 1000 a=-\sin 999 a$, a total of 500 equalities. From this, we obtain that $\sin 2 a \cdot \sin 4 a \cdot \sin 6 a \cdot \ldots \cdot \sin 1998 a=\sin a \cdot \sin 2 a \cdot \sin 3 a \cdot \ldots \cdot \sin 999 a$. None of these sines are zero, so $P=2^{-999}$.

![](https://cdn.mathpix.com/cropped/2024_06_04_7a3ed3d68324df2727b5g-15.jpg?height=1042&width=953&top_left_y=84&top_left_x=563)

Fig. 5: to the solution of problem 3.2