how do I show this :$\sum_{k=0}^{2n }\binom{2n}{k} \sin ((n-k)x)=0$ , for every real $x$ and for every integer $n$? My attempt fails to show this formula $\sum_{k=0}^{2n }\binom{2n}{k} \sin ((n-k)x)=0$ which I have accrossed in my textbook, using induction proof, but I think by induction seems very hard, I want to know if there is any simple way to show the titled identity that is true for every for every real $x$ and for every integer $n$ ? Thanks in advance. 
 A: Hint: What is $$\binom{2n}k\sin((n-k)x)+\binom{2n}{2n-k}\sin((n-(2n-k))x)?$$
A: Just to expand the MathematicalPhysicist's idea, we have $$\sum_{k=0}^{2n}\dbinom{2n}{k}\sin\left(\left(n-k\right)x\right)=\mathrm{Im}\left(\sum_{k=0}^{2n}\dbinom{2n}{k}e^{i\left(n-k\right)x}\right)$$ and, by the binomial theorem, we get $$\sum_{k=0}^{2n}\dbinom{2n}{k}e^{i\left(n-k\right)x}=e^{nix}\sum_{k=0}^{2n}\dbinom{2n}{k}e^{-kxi}=e^{nix}\left(1+e^{-ix}\right)^{2n}$$ $$=\left(e^{ix/2}+e^{-ix/2}\right)^{2n}=4^{n}\cos^{2n}\left(\frac{x}{2}\right)$$ so $$\sum_{k=0}^{2n}\dbinom{2n}{k}\sin\left(\left(n-k\right)x\right)=\color{red}{0}.$$
A: Hint:  use the symmetry of Pascal's triangle.  To make it obvious, write $j=k-n$ and take $j$ from $-n$ to $+n$. 
A: Note that $\binom{2n}{n} \sin ((n-n)x) = 0$. Now go $i = 1,\ldots , n$ steps both to the left and to the right within the given sum and compare the terms:
$\sum_{k=0}^{2n }\binom{2n}{k} \sin ((n-k)x)= \sum_{i=1}^n\binom{2n}{n-i} \sin ((n-(n-i))x) + \sum_{i=1}^n\binom{2n}{n+i} \sin ((n-(n+i))x) \stackrel{\binom{2n}{n-i}= \binom{2n}{n+i}}{=} \sum_{i=1}^n\binom{2n}{n-i} \sin (ix) - \sum_{i=1}^n\binom{2n}{n-i} \sin (ix) = 0$
