Summation of series with binomial coefficients

Question

The value of $$\sum {n\choose n-r} (n-r) \sin(r\cdot \pi/n)$$ where $r\in (0 ..,n)$ is equal to? I think the question can be solved by writing the series in reverse order but I am not able to solve further.

Answer 1

HINT:

$$(n-r)\cdot\dfrac{n!}{r!\cdot(n-r)!}=n\dfrac{n-1}{r!\cdot\{(n-1)-r\}!}=n\binom{n-1}r$$

Now $\sin\dfrac{r\pi}n=$ imaginary of $e^{r\pi i/n}$

$$\sum_{r=0}^n(n-r)\cdot\dfrac{n!}{r!\cdot(n-r)!}e^{r\pi i/n}=n\sum_{r=0}^n\binom{n-1}r(e^{\pi i/n})^r=n(1+e^{\pi i/n})^{n-1}$$

Finally, $1+e^{2iy}=e^{iy}(e^{iy}+e^{-iy})=2e^{iy}\cos y$