326. Simplify

$$
\begin{aligned}
\cos \alpha+C_{n}^{1} \cos 2 \alpha+C_{n}^{2} \cos 3 \alpha+\ldots+C_{n}^{n-1} & \cos n \alpha+ \\
& +\cos (n+1) \alpha
\end{aligned}
$$

and

$\sin \alpha+C_{n}^{1} \sin 2 \alpha+C_{n}^{2} \sin 3 \alpha+\ldots+C_{n}^{n-1} \sin n \alpha+$ $+\sin (n+1) \alpha$
326. It is required to compute the real part and the coefficient of the imaginary part of the sum

$(\cos \alpha+i \sin \alpha)+C_{n}^{1}(\cos 2 \alpha+i \sin 2 \alpha)+$

$$
+C_{n}^{2}(\cos 3 \alpha+i \sin 3 \alpha)+\ldots+(\cos (n+1) \alpha+i \sin (n+1) \alpha)
$$

Denoting $\cos \alpha+i \sin \alpha$ by $x$ and using de Moivre's formula and the binomial theorem, we transform this sum as follows:

$$
\begin{aligned}
x+C_{n}^{1} x^{2} & +C_{n}^{2} x^{3}+\ldots+x^{n+1}=x(x+1)^{n}= \\
& =(\cos \alpha+i \sin \alpha)(\cos \alpha+1+i \sin \alpha)^{n}= \\
& =(\cos \alpha+i \sin \alpha)\left(2 \cos ^{2} \frac{\alpha}{2}+2 i \cos \frac{\alpha}{2} \sin \frac{\alpha}{2}\right)^{n}= \\
& =2^{n} \cos ^{n} \frac{\alpha}{2}(\cos \alpha+i \sin \alpha)\left(\cos \frac{n \alpha}{2}+i \sin \frac{n \alpha}{2}\right)= \\
& =2^{n} \cos ^{n} \frac{\alpha}{2}\left(\cos \frac{n+2}{2} \alpha+i \sin \frac{n+2}{2} \alpha\right)
\end{aligned}
$$

From this, it follows that

$\cos \alpha+C_{n}^{1} \cos 2 \alpha+C_{n}^{2} \cos 3 \alpha+\ldots+\cos (n+1) \alpha=$

$$
=2^{n} \cos ^{n} \frac{\alpha}{2} \cos \frac{n+2}{2} \alpha
$$

$\sin \alpha+C_{n}^{1} \sin 2 \alpha+C_{n}^{2} \sin 3 \alpha+\ldots+\sin (n+1) \alpha=$

$$
=2^{n} \cos ^{n} \frac{\alpha}{2} \sin \frac{n+2}{2} \alpha
$$