when I study digital signal processing course got this following math Problem (finite impulse response):the book give this reslut(But I can't see the proof on the book):
Question: Give the postive integer $N>2$, if for any real numbers $x$,we have $$\sum_{n=0}^{N-1}a(n)\sin{[(\tau-n)x]}=0$$ if and only if $$\begin{cases} \tau=\dfrac{N-1}{2}\\ a(n)=a(N-1-n),0\le n\le N-1 \end{cases}$$
Now I can prove $\Longleftarrow $
becasue if $$\tau=\dfrac{N-1}{2},~~~ a(n)=a(N-1-n)$$ then $$\begin{align*}\sum_{n=0}^{N-1}a(n)\sin{[(\tau-n)x]}&=\sum_{n=0}^{N-1}a(n)\sin{[(\frac{N-1}{2}-n)x]}\\ &=\sum_{n=0}^{\frac{N-1}{2}}a(n)\sin{[(\frac{N-1}{2}-n)x]}+\sum_{n=\frac{N+1}{2}}^{N-1}a(N-1-n)\sin{[(\frac{N-1}{2}-n)x]}\\ &=\sum_{n=0}^{\frac{N-1}{2}}a(n)\sin{[(\frac{N-1}{2}-n)x]}-\sum_{n=0}^{\frac{N-1}{2}}a(n)\sin{[(\frac{N-1}{2}-n)x]}\\ &=0\end{align*}$$
But How to prove $\Longrightarrow$
This