Bổ đề: Cho $(n,a,b)\in \mathbb{N}\times \mathbb{R}\times\mathbb{R}$. Tính các tổng:
$C_n=\sum_{k=0}^{n} \cos(a+kb),S_n=\sum_{k=0}^{n} \sin(a+kb)$
Chứng minh bổ đề:
Xét $\displaystyle C_n+iS_n=\sum_{k=0}^n e^{i(a+bk)}=e^{ia}\sum_{k=0}^n (e^{ib})^k$
*) Nếu $b\in 2\pi \mathbb{Z}$ thì $C_n=(n+1)\cos a, \, S_n=(n+1)\sin a.$
*) Nếu $b\notin 2\pi \mathbb{Z}$ thì:
$\displaystyle C_n+iS_n=e^{ia}\frac{(e^{ib})^{n+1}-1}{e^{ib}-1}=e^{ia}\frac{\displaystyle{ e^{i\frac{(n+1)b}{2}}2i\sin \frac{n+1}{2}b}}{e^{i\frac{b}{2}}2i\sin \displaystyle{\frac{b}{2}}}=e^{i(a+\frac{nb}{2})}\frac{\displaystyle{\sin\frac{n+1}{2}b}}{\displaystyle{\sin\frac{b}{2}}}$
Suy ra $\displaystyle C_n=\cos(a+\frac{nb}{2})\frac{\displaystyle{\sin\frac{n+1}{2}b}}{\displaystyle{\sin\frac{b}{2}}},\,S_n=\sin(a+\frac{nb}{2})\frac{\displaystyle{\sin\frac{n+1}{2}b}}{\displaystyle{\sin\frac{b}{2}}}$
Quay lại bài toán, ta có:
Vì $\sin0=0$ và $|\sin k|\leq 1$ nên:
$\displaystyle \sum_{k=1}^n |\sin k|=\sum_{k=0}^n |\sin k| \geq \sum_{k=0}^n \sin^2 k=\frac{1}{2}\sum_{k=0}^n (1-\cos 2k)$
$\displaystyle \qquad \qquad \;\;\,=\frac{n+1}{2}-\frac{1}{2}\sum_{k=0}^n \cos 2k=\frac{n+1}{2}-\frac{1}{2}\frac{\sin (n+1)}{\sin 1}\cos n$
Vì $\displaystyle |\frac{\sin (n+1)}{\sin 1}\cos n| \leq \frac{1}{\sin 1}$
nên $\displaystyle \sum_{k=1}^{n}|\sin k|\geq \frac{n+1}{2}-\frac{1}{2\sin 1}$