331. Simplify the products:

a) $\sin \frac{\pi}{2 n+1} \sin \frac{2 \pi}{2 n+1} \sin \frac{3 \pi}{2 n+1} \ldots \sin \frac{n \pi}{2 n+1}$

and

$$
\sin \frac{\pi}{2 n} \sin \frac{2 \pi}{2 n} \sin \frac{3 \pi}{2 n} \ldots \sin \frac{(n-1) \pi}{2 n}
$$

b) $\cos \frac{\pi}{2 n+1} \cos \frac{2 \pi}{2 n+1} \cos \frac{3 \pi}{2 n+1} \ldots \cos \frac{n \pi}{2 n+1}$

II

$$
\cos \frac{\pi}{2 n} \cos \frac{2 \pi}{2 n} \cos \frac{3 \pi}{2 n} \ldots \cos \frac{(n-1) \pi}{2 n}
$$
331. a) First solution. The numbers $\sin ^{2} \frac{\pi}{2 n+1}, \quad \sin ^{2} \frac{2 \pi}{2 n+1}, \cdots$ $\ldots \sin ^{2} \frac{n_{\pi}}{2 n+1}$ are the roots of the equation of degree $n$, obtained in the solution of problem 329a). The coefficient of the leading term $x^{n}$ of this equation is equal to $(-1)^{n}\left[C_{2 n+1}^{1}+C_{2 n+1}^{3}+\ldots+C_{2 n+1}^{2 n-1}+1\right]$. But the sum in the square brackets is half of the total sum of binomial coefficients $1+C_{2 n+1}^{1}+C_{2 n+1}^{2}+\ldots+C_{2 n+1}^{2 n}+1$, which is known to be $(1+1)^{2 n+1}=2^{2 n+1}$; therefore, the coefficient of $x^{n}$ in our equation is equal to $(-1)^{n} 2^{2 n}$. Further, the constant term of this equation is $C_{2 n+1}^{1}=2 n+1$. But the product of the roots of an equation of degree $n$ is equal to the constant term of the reduced equation (an equation with the coefficient of the leading term equal to one), multiplied by $(-1)^{n}$, i.e., it is equal to the constant term, multiplied by $(-1)^{n}$ and divided by the coefficient of the leading term. Hence we have:

$$
(-1)^{n} \sin ^{2} \frac{\pi}{2 n+1} \sin ^{2} \frac{2 \pi}{2 n+1} \ldots \sin ^{2} \frac{n \pi}{2 n+1}=(-1)^{n} \frac{2 n+1}{2^{2 n}}
$$

and, therefore,

$$
\sin \frac{\pi}{2 n+1} \sin \frac{2 \pi}{2 n+1} \cdots \sin \frac{n \pi}{2 n+1}=\frac{\sqrt{2 n+1}}{2^{n}}
$$

## Similarly, it can be proven that

$$
\sin \frac{\pi}{2 n} \sin \frac{2 \pi}{2 n} \ldots \sin \frac{(n-1) \pi}{2 n}=\frac{\sqrt{n}}{2^{n-1}}
$$

Second solution. The roots of the equation $x^{2 n}-1=0$ are

$$
\begin{gathered}
1,-1, \quad \cos \frac{\pi}{n}+i \sin \frac{\pi}{n}, \quad \cos \frac{2 \pi}{n}+i \sin \frac{2 \pi}{n} \\
\cos \frac{3 \pi}{n}+i \sin \frac{3 \pi}{n}, \ldots, \cos \frac{(2 n-1) \pi}{n}+i \sin \frac{(2 n-1) \pi}{n}
\end{gathered}
$$

## Therefore, we have:

$$
\begin{array}{r}
x^{2 n}-1=(x-1)(x+1)\left(x-\cos \frac{\pi}{n}-i \sin \frac{\pi}{n}\right) \times \\
\times\left(x-\cos \frac{2 \pi}{n}-i \sin \frac{2 \pi}{n}\right) \ldots\left(x-\cos \frac{(n-1) \pi}{n}-i \sin \frac{(n-1) \pi}{n}\right) \times \\
\times\left(x-\cos \frac{(n+1) \pi}{n}-i \sin \frac{(n+1) \pi}{n}\right) \times \ldots \\
\ldots \times\left(x-\cos \frac{(2 n-1) \pi}{n}-i \sin \frac{(2 n-1) \pi}{n}\right)
\end{array}
$$

But $\cos \frac{(2 n-k) \pi}{n}=\cos \frac{k \pi}{n}, \sin \frac{(2 n-k) \pi}{n}=-\underline{\sin } \frac{k \pi}{n}$; hence it follows that

$$
\begin{aligned}
& \begin{aligned}
& \text { follows that } \\
&\left(x-\cos \frac{\pi}{n}-i \sin \frac{\pi}{n}\right)\left(x-\cos \frac{(2 n-1) \pi}{n}-i \sin \frac{(2 n-1) \pi}{n}\right)= \\
&=x^{2}-2 x \cos \frac{\pi}{n}+1, \\
& \begin{aligned}
&\left(x-\cos \frac{2 \pi}{n}-i \sin \frac{2 \pi}{n}\right)\left(x-\cos \frac{(2 n-2) \pi}{n}-i \sin \frac{(2 n-2) \pi}{n}\right)= \\
&=x^{2}-2 x \cos \frac{2 \pi}{n}+1,
\end{aligned} \\
& \begin{aligned}
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
\left(x-\cos \frac{(n-1) \pi}{n}-i \sin \frac{(n-1) \pi}{n}\right) \times
\end{aligned} \\
& \quad \times\left(x-\cos \frac{(n+1) \pi}{n}-i \sin \frac{(n+1) \pi}{n}\right)=x^{2}-2 x \cos \frac{(n-1) \pi}{n}+1 .
\end{aligned}
\end{aligned}
$$

Therefore, the factorization of the polynomial $x^{2 n}-1$ can be rewritten as:

$$
\begin{aligned}
x^{2 n}-1=\left(x^{2}-1\right