The Fejer kernel has this $\sin$ closed form.

We can prove the equality for the Fejér kernel in the following way. Using the formula for the geometric progression and the fact that $e^{i\theta}-e^{-i\theta}=2i\sin\theta$ for each $\theta\in\mathbb R$, \begin{align*} D_k(x) &=e^{-ikx}\sum_{s=0}^{2k}e^{isx}\\ &=e^{-ikx}\frac{1-e^{ix(2k+1)}}{1-e^{ix}}\\ &=\frac{e^{-ix(k+1/2)}-e^{ix(k+1/2)}}{e^{-ix/2}-e^{ix/2}}\\ &=\frac{\sin(x(k+1/2))}{\sin(x/2)}. \end{align*} Using the product-to-sum identity and the power reduction formula, \begin{align*} F_n(x) &=\frac1{n\sin(x/2)}\sum_{k=0}^{n-1}\sin(x(k+1/2))\\ &=\frac1{2n\sin^2(x/2)}\sum_{k=0}^{n-1}2\sin(x(k+1/2))\sin(x/2)\\ &=\frac1{2n\sin^2(x/2)}\sum_{k=0}^{n-1}[\cos(kx)-\cos((k+1)x)]\\ &=\frac{1-\cos (nx)}{2n\sin^2(x/2)}\\ &=\frac1n\biggl[\frac{\sin(nx/2)}{\sin(x/2)}\biggr]^2. \end{align*} The trigonometric identities can be found here.

Remember that $NF_N(x)=D_0(x)+\dots+D_{N-1}(x)$ where $D_n(x)$ is the Dirichlet kernel. Therefore , if $\omega = e^{ix}$ we have $$NF_N(x)=\sum_{n=0}^{N-1}\frac{\omega^{-n}-\omega^{n+1}}{1-\omega}$$ Then we have $$\sum_{n=0}^{N-1}\omega^{-n}=1+1/\omega+\dots+1/\omega^{N-1}=\frac{1-\omega^{-N}}{1-\omega^{-1}}=\omega\frac{\omega^{-N}-1}{1-\omega}$$ and $$\sum_{n=0}^{N-1}\omega^{n+1}=\omega(1+\omega+\dots+\omega^{N-1})=\omega\frac{1-\omega^N}{1-\omega}$$ Thus $$NF_N(x)=\frac{\sum_{n=0}^{N-1}\omega^{-n}-\sum_{n=0}^{N-1}\omega^{n+1}}{1-\omega} =\omega\frac{\omega^N-2+\omega^{-N}}{(1-\omega)^2}=\frac{(\omega^{N/2}-\omega^{-N/2})^2}{(\omega^{1/2}-\omega^{-1/2})^2}=\frac{\sin^2(Nx/2)}{\sin^2(x/2)}$$ So have get $$F_N(x)=\frac{1}{N}\frac{\sin^2(Nx/2)}{\sin^2(x/2)}$$