How do I show that the sequence of functions: $$ f_n(x):=\begin{cases} \biggl| {{\cos (x/2)-\cos \bigl(\,(n+{1\over2})x\,\bigr)}\over 2\sin(x/2) } \biggr|, & ~ x \neq 2k\pi,~k\in\mathbb{Z}\\ 0, &~x=2k\pi,~k\in\mathbb{Z} \end{cases} $$ where $x\in\mathbb{R}$, doesn't converge uniformly?
My first attempt was to construct a sequence like $x_n:=\frac{\pi}{2n}$, and then show that $\Vert f_n-f_m\Vert_{\infty}\geq| f_n(x_n)-f_m(x_n)|>C$ for a given constant $C>0$. But the resulting expressions I got only made it worse. So I am not sure if this is the way to go?
Uniform convergence implies pointwise convergence everywhere, so we only need to show that it doesn't converge pointwisely, and this in a single point.
We'll show that the sequence $(a_n)_n$ = $(f_n(\pi/2))_n$ has no limit. Indeed, $(a_n)_n = (|\sin (n \pi/4) \sin ((n+1) \pi/4) / \sin(\pi/4)|)_n$ is just the sequence $1, 1, 0, 0, 1, 1, 0, 0, \ldots, $ and this sequence doesn't converge.