I have Fejer's kernel, $F_n(\alpha):=\frac{1}{2(n+1)}\frac{\sin^2((n+1)\alpha/2)}{\sin^2(\alpha/2)}$; I also have $\sigma_n(f,x):= 1/\pi \int_{-\pi}^{\pi} f(t)F_n(x-t)dt$.
I want to show that for $t_k:= \cos(kx)$ and for $t_k^* := \sin(kx)$ we have the following identity: $$\sigma_n(t_k) = \frac{n-k-1}{n+1}t_k$$ The same above last identity also with $t_k^*$ replacing $t_k$.
How to do this computation? (I mean the denominator seems to be intimidating) Thanks.
Definitions:
It seems best to use the exponential basis. Recall that the functions $\{ e^{ik x} \}_{k \in \mathbb{Z}}$ form an orthonormal system w.r.t. the inner product $$\langle f,g \rangle = \frac{1}{2\pi} \int_{-\pi}^{\pi} f \overline{g} dx$$ in $L_2([-\pi,\pi])$.
For any natural $n$, we have the Dirichlet kernel $$D_n(x) = \sum_{s=-n}^{n} e^{is x},$$ which has the following (defining) property $$(D_n*f)(x)=\frac{1}{2\pi}\int_{-\pi}^\pi f(y)D_n(x-y)\,dy=\sum_{k=-n}^n \hat{f}(k)e^{ikx}.$$
The Fejér kernel is defined as some sort of smoothing of the Dirichlet kernel, namely $$F_{n}(x)={\frac {1}{n}}\sum _{{k=0}}^{{n-1}}D_{k}(x).$$ It can also be expressed as $$F_{n}(x)={\frac {1}{n}}\left({\frac {\sin {\frac {nx}{2}}}{\sin {\frac {x}{2}}}}\right)^{2}.$$ (so my conventions slightly differ from yours. I will stick to mine.)
Your $\sigma_n(f,x)$ is just $f * F_n$. Combining the above, we see that this convolution may be expressed as $$(*) f*F_{n}={\frac {1}{n}}\sum _{{k=0}}^{{n-1}}f*D_{k}.$$
When I plug $f=f_k=\cos(kx) = \frac{e^{ik x} + e^{-ikx}}{2}$ ($k$ - non-negative integer), I get
$$(**)f_k*D_{n}= \begin{cases} 0 & n<k \\ f_k & n \ge k \end{cases}.$$
Similarly, if I plug $f=f_k=\sin(kx) = \frac{e^{ik x} - e^{-ikx}}{2i}$, I get
$$(***) f_k*D_{n}= \begin{cases} 0 & n<k \\ f_k & n \ge k \end{cases}.$$
Plugging $f_k=\sin(kx), \cos(kx)$ in $(*)$ and using $(**),(***)$, we see
$$f_k*F_{n}={\frac {1}{n}}\sum _{{n-1 \ge j \ge k}} f_k = \begin{cases} 0 & n \le k \\ \frac{n-k}{n} f_k & n > k \end{cases},$$ as needed.