高端三角恒等变换（sin）
由正弦的降幂公式：
sin^{2r}{x} = \frac{1}{2r-1} \sum_{k=0}^{r-1}(-1)^{r+k}(^{2r}_{k})cos(2r-rk)x+\frac{1}{4^{r}}(_{r}^{2r})
可得：
sin^{2r}\frac{mπ}{2n+1}=\frac{1}{2^{2r-1}} \sum_{k=0}^{r-1}(-1)^{r+k}(^{2r}_{k})cos(2r-rk)\frac{mπ}{2n+1} +\frac{1}{4^{r}}(_{r}^{2r})
sin^{2r}\frac{mπ}{2n}=\frac{1}{2^{2r-1}} \sum_{k=0}^{r-1}(-1)^{r+k}(^{2r}_{k})cos(2r-rk)\frac{mπ}{2n} +\frac{1}{4^{r}}(_{r}^{2r})
又由余弦的基本等差求和公式：
\sum^{n}_{m=1} \cos2mx=\frac{\sin(2n+1)x}{2\sin x}-\frac{1}{2}
\sum^{n-1}_{m=1} \cos2mx=\frac{\sin(2n-1)x}{2\sin x}-\frac{1}{2}
可得：
\sum^{n}_{m=1} \cos[(2r-2k)\frac{mπ}{2n}]=\frac{\sin \frac{(2r-2k)π}{2}}{2\sin \frac{(2r-2k)π}{2(2n+1)}}-\frac{1}{2}=-\frac{1}{2}
\sum^{n-1}_{m=1} \cos[(2r-2k)\frac{mπ}{2n+1}]=\frac{\sin \frac{(2r-2k)(2n-1)π}{4n}}{2\sin \frac{(2r-2k)π}{4n}}-\frac{1}{2}=\frac{1}{2}((-1)^{r-k+1}-1)
所以:
\sum^{n}_{m=1}\sin^{2r}\frac{mπ}{2n+1}=\frac{1}{4r}[\sum^{r-1}_{k=0}(-1)^{r+k+!}(^{2r}_{k})+(^{2r}_r)n]
\sum^{n}_{m=1}\sin^{2r}\frac{mπ}{2n}=\frac{1}{4r}[\sum^{r-1}_{k=0}(-1)^{r+k}(^{2r}_{k})((-1)^{r-k+1}-1)+(^{2r}_r)(n-1)]
进一步化简：
利用公式(^n_k)=(^{n-1}_k)+(^{n-1}_{k-1})可得
\sum^{r-1}_{k=0}(-1)^{r+k+1}(^{2r}_k)
=(-1)^{r+1}+\sum^{r-1}_{k=1}(-1)^{r+1+k}[(^{2r-1}_k)+(^{2r-1}_{k-1})]
=(-1)^{r+1}+\sum^{r-1}_{k=1}(-1)^{r+1+k}[(^{2r-1}_k)-(-1)^{r+1+k-1}(^{2r-1}_{k-1})]
=(-1)^{r-1}-(-1)^{r-1}+(^{2r-1}_{r-1})=\frac{(2r-1)!}{(r-1)!r!}]
=\frac{r!(2r-1)!}{(r-1)!(2r)!}·\frac{(2r)!}{(r!)^2}=\frac{1}{2}(^{2r}_r)
又有
\sum^{r-1}_{k=0}(^{2r}_k)=\frac{1}{2}[\sum^{r-1}_{k=0}(^{2r}_k)+\sum^{2r}_{k=r+1}(^{2r}_k)]
=\frac{1}{2}[\sum^{2r}_{k=0}(^{2r}_k)-(^{2r}_r)]
=\frac{1}{2}4^r(^{2r}_r)n-\frac{1}{2}
所以
\sum^{n}_{m=1}\sin^{2r}\frac{mπ}{2n+1}=\frac{1}{4^r}(^{2r}_r)(n+\frac{1}{2})
\sum^{n-1}_{m=1}\sin^{2r}\frac{mπ}{2n}=\frac{1}{4^r}(^{2r}_r)n-\frac{1}{2}
（随便写写玩玩的）