Tricky Sum involving Binomial Coefficients and Sine I am stumped by the sum
$$\sum_{x=0}^n \binom{n}{x}\sin\big(\frac{\pi x}{n}\big)$$
but I can't figure it out. I tried expanding the taylor series of sine and using Euler's identity, but to no avail. Any hints?
PLEASE do not give me a full solution - I just need a hint.
Thanks!
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{x = 0}^{n}{n \choose x}\sin\pars{\pi x \over n} & =
\Im\sum_{x = 0}^{n}{n \choose x}\expo{\ic\pi x/n} =
\Im\sum_{x = 0}^{n}{n \choose x}\pars{\expo{\ic\pi/n}}^{x} =
\Im\pars{1 + \expo{\ic\pi/n}}^{n}
\\[5mm] & =
\Im\bracks{1 + \cos\pars{\pi \over n} + \ic\sin\pars{\pi \over n}}^{\,n}
\\[5mm] & =
\Im\bracks{\root{2 + 2\cos\pars{\pi \over n}}
\exp\pars{\ic\arctan\pars{\sin\pars{\pi/n} \over 1 + \cos\pars{\pi/n}}}}^{\,n}
\\[5mm] & =
2^{n/2}\,\bracks{2\cos^{2}\pars{\pi \over 2n}}^{\,n/2}\
\sin\pars{n\arctan\pars{2\sin\pars{\pi/\bracks{2n}}\cos\pars{\pi/\bracks{2n}} \over 2\cos^{2}\pars{\pi/\bracks{2n}}}}
\\[5mm] & =
2^{n}\,\verts{\cos\pars{\pi \over 2n}}^{\,n}
\sin\pars{n\arctan\pars{\tan\pars{\pi \over 2n}}}
\\[5mm] & =
2^{n}\,\verts{\cos\pars{\pi \over 2n}}^{\,n}\sin\pars{\pi \over 2} =
\bbx{2^{n}\,\verts{\cos\pars{\pi \over 2n}}^{\,n}} \\ &
\end{align}
