Calculate:

a) $\cos 2 \pi / 7 + \cos 4 \pi / 7 + \cos 6 \pi / 7$

b) $\cos 2 \pi / 7 \cos 4 \pi / 7 \cos 6 \pi / 7$.
a) The first method. Let $z=\cos 2 \pi / 7 + i \sin 2 \pi / 7$. Then $-1=z+z^{2}+z^{3}+z^{4}+z^{5}+z^{6}=z+z^{2}+z^{3}+\overline{z+z^{2}+z^{3}}=$ $2 \operatorname{Re}\left(z+z^{2}+z^{3}\right)=2\left(\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}\right)$

The second method. $2 \sin \alpha(\cos 2 \alpha+\cos 4 \alpha+\cos 6 \alpha)=\sin 3 \alpha-\sin \alpha+\sin 5 \alpha-\sin 3 \alpha+\sin 7 \alpha-\sin 5 \alpha=\sin 7 \alpha-\sin \alpha$. b) $\cos \alpha \cos 2 \alpha \cos 3 \alpha=1 / 2 \cos 2 \alpha(\cos 2 \alpha+\cos 4 \alpha)=1 / 4(1+\cos 4 \alpha+\cos 2 \alpha+\cos 6 \alpha)$.

## Answer

a) $-1 / 2$;
b) $1 / 8$.

It is known that $z+z^{-1}=2 \cos \alpha$.

a) Prove that $z^{n}+z^{-n}=2 \cos n \alpha$.

b) How is $z^{n}+z^{-n}$ expressed in terms of $y=z+z^{-1}$?

## Hint

Transition to conjugate numbers in the equality $z+z^{-1}=2 \cos \alpha$ and compute $z$.

## Solution

a) If $z+z^{-1}=2 \cos \alpha$, then $\bar{z}+\bar{z}^{-1}=2 \cos \alpha$ as well. Subtracting, we get $\frac{(z-\bar{z})(z \bar{z}-1)}{z \bar{z}}=0$. Therefore, either $z=\bar{z}$, meaning $z$ is a real number, or $|z|=1$. The first case is possible only when $z= \pm 1$ (otherwise $\left|z+z^{-1}\right|>2 \geq 2 \cos \alpha$). In the second case, $\bar{z}=z^{-1}$,

which means $\operatorname{Re} z=\cos \alpha$. Hence, $z=\cos \alpha \pm i \sin \alpha$, and by De Moivre's formula, $z^{n}+z^{-n}=2 \cos n \alpha$.

b) $z^{n}+z^{-n}=(\cos \alpha+i \sin \alpha)^{n}+(\cos \alpha-i \sin \alpha)^{n}=$

$$
\begin{gathered}
=2\left(\cos ^{n} \alpha-C_{n}^{2} \cos ^{n-2} \alpha \sin ^{2} \alpha+C_{n}^{4} \cos ^{n-4} \alpha \sin ^{4} \alpha-\ldots\right)= \\
2^{1-n}\left(y^{n}-C_{n}^{2} y^{n-2}\left(4-y^{2}\right)+C_{n}^{4} y^{n-4}\left(4-y^{2}\right)^{2}-\ldots\right)
\end{gathered}
$$

## Answer

b) $z^{n}+z^{-n}=2^{1-n}\left(y^{n}-C_{n}^{2} y^{n-2}\left(4-y^{2}\right)+C_{n}^{4} y^{n-4}\left(4-y^{2}\right)^{2}-\ldots\right)$.