23.7. Using the result of problem 23.6, calculate the following sums and products:

a) $\operatorname{ctg}^{2} \frac{\pi}{2 n+1}+\operatorname{ctg}^{2} \frac{2 \pi}{2 n+1}+\ldots+\operatorname{ctg}^{2} \frac{n \pi}{2 n+1}$;

b) $\frac{1}{\sin ^{2} \frac{\pi}{2 n+1}}+\frac{1}{\sin ^{2} \frac{2 \pi}{2 n+1}}+\ldots+\frac{1}{\sin ^{2} \frac{n \pi}{2 n+1}}$;

c) $\sin \frac{\pi}{2 n+1} \sin \frac{2 \pi}{2 n+1} \ldots \sin \frac{n \pi}{2 n+1}$.

## 23.3. Roots of Unity

A root of the $n$-th degree from unity is a complex number $\varepsilon$ for which $\varepsilon^{n}=1$. A primitive root of the $n$-th degree from unity is a complex number $\varepsilon$ for which $\varepsilon^{n}=1$ and $\varepsilon^{k} \neq 1$ for $k=1,2, \ldots, n-1$.
23.7. а) The considered sum is equal to

$$
\frac{C_{2 n+1}^{3}}{C_{2 n+1}^{1}}=\frac{n(2 n-1)}{3}
$$

б) The identity $\frac{1}{\sin ^{2} \alpha}=1+\operatorname{ctg}^{2} \alpha$ shows that the considered sum is equal to $\frac{n(2 n-1)}{3}+n=\frac{2 n(n+1)}{3}$.

в) Answer: $\frac{\sqrt{2 n+1}}{2^{n}}$. The square of the considered product, up to the sign, is equal to the ratio of the free term of the polynomial from problem 23.6 а) to the coefficient of $x^{n}$. The free term is equal to $C_{2 n+1}^{1}$; the coefficient of $x^{n}$, up to the sign, is equal to $C_{2 n+1}^{1}+C_{2 n+1}^{3}+\ldots+C_{2 n+1}^{2 n+1}$. If we add $C_{2 n+1}^{0}+C_{2 n+1}^{2}+\ldots+C_{2 n+1}^{2 n}$ to the last sum, we get $2^{2 n+1}$. But both these sums are equal, since $C_{2 n+1}^{2 k+1}=C_{2 n+1}^{2 n-2 k}$. Therefore, the square of the considered product is equal to $\frac{2 n+1}{2^{2 n}}$. It is also clear that the considered product is positive.