### 3.493 Find the sum $1+\cos 4 \alpha+\cos 8 \alpha+\ldots+\cos 4 n \alpha$.
## Solution.

$2 \sin 2 \alpha(1+\cos 4 \alpha+\cos 8 \alpha+\ldots+\cos 4 n \alpha)=2 \sin 2 \alpha+2 \sin 2 \alpha \cdot \cos 4 \alpha+\ldots+$ $+2 \sin 2 \alpha \cdot \cos 4 n \alpha=2 \sin 2 \alpha+(\sin 6 \alpha-\sin 2 \alpha)+(\sin 10 \alpha-\sin 6 \alpha)+\ldots+$ $+(\sin (4 n \alpha-2 \alpha)-\sin (4 n \alpha-6 \alpha))+(\sin (4 n \alpha+2 \alpha)-\sin (4 n \alpha-2 \alpha))=\sin 2 \alpha+$ $+\sin (4 n \alpha+2 \alpha)=2 \sin 2 \alpha(n+1) \cdot \cos 2 n \alpha \Rightarrow 1+\cos 4 \alpha+\ldots+\cos 4 n \alpha=$ $=\frac{\sin 2 \alpha(n+1) \cdot \cos 2 n \alpha}{\sin 2 \alpha}$.