Property of Fejer kernel Let
$$
 F_n(x) = \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2
$$
be the n-th Fejer-Kernel. Then 
$$
 \forall \epsilon > 0, r < \pi : \exists N \in \mathbb{N} : \forall n \ge N : \int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t < \epsilon
$$
The proof of this goes like this
\begin{align*}
 \int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t 
 & = \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2 \, \mathrm{d} t \\
 & \le \frac{1}{n} \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{\sin^2(\frac{1}{2}t)} \, \mathrm{d} t \\
 & \le \frac{1}{n} \int_{[-\pi,\pi]} \frac{1}{\sin^2(\frac{1}{2}r)} \, \mathrm{d} t  \qquad (*) \\
& = \frac{2\pi}{n} \frac{1}{\sin^2(\frac{1}{2}r)} \\
& < \epsilon \quad \textrm{for certain } ~ n \ge N.
\end{align*}
The step i marked with (*) is totally unclear to me, why could there replaced $t$ by $r$, because it need not to be the case that $\sin^2(0.5*r) < \sin^2(0.5*t)$ as i think?

Note that you're integrating over the set $r < |x| \le \pi$.  
Let $f(x) = \sin^2(\frac{1}{2}x)$.  Differentiating, we get that $f'(x) = \frac{\sin(x)}{2}$.  Now if $x \in (0,\pi]$, then $\sin(x) > 0$, and so $f$ is increasing on $(0,\pi]$.  Thus $\sin^2(\frac{1}{2} r) < \sin^2(\frac{1}{2}x)$ for $0 < r < x \le \pi$.  Similarly, we see that since $\sin(x) < 0$ for $x \in [-\pi, 0)$, so $f$ is decreasing on this interval.  So if you pick some $-\pi \ge x > -r > 0$, it follows that $\sin^2(\frac{1}{2}r) < \sin^2(\frac{1}{2}x)$, which gives the desired inequality.
It may be somewhat instructive to look at a graph
You can easily see that the inequality should hold over the desired interval.  The idea of checking this using calculus is a standard tool for these sorts of estimates.
