一组三角求和

lemondian

三角求和：
已知$r\inN^*$,求：
1.$\sum_{k=1}^n\sin^{2r}\dfrac{k\pi}{2n+1}$;

2.$\sum_{k=1}^{n-1}\sin^{2r}\dfrac{k\pi}{2n}$;

3.$\sum_{k=1}^n\cos^{2r}\dfrac{k\pi}{2n+1}$;

4.$\sum_{k=1}^{n-1}\cos^{2r}\dfrac{k\pi}{2n}$.

战巡

有这么两个公式：
\[\sin^{2r}(\theta)=\frac{C_{2r}^r}{2^{2r}}+\frac{1}{2^{2r-1}}\sum_{j=0}^{r-1}(-1)^{r-j}C_{2r}^j\cos((2r-2j)\theta)\]
以及
\[\cos^{2r}(\theta)=\frac{C_{2r}^r}{2^{2r}}+\frac{1}{2^{2r-1}}\sum_{j=0}^{r-1}C_{2r}^j\cos((2r-2j)\theta)\]

下面只示范一个
\[\sin^{2r}(\frac{k\pi}{2n+1})=\frac{C_{2r}^r}{2^{2r}}+\frac{1}{2^{2r-1}}\sum_{j=0}^{r-1}(-1)^{r-j}C_{2r}^j\cos((2r-2j)\frac{k\pi}{2n+1})\]
\[\sum_{k=1}^n\sin^{2r}(\frac{k\pi}{2n+1})=\frac{nC_{2r}^r}{2^{2r}}+\frac{1}{2^{2r-1}}\sum_{j=0}^{r-1}(-1)^{r-j}C_{2r}^j\sum_{k=1}^n\cos((2r-2j)\frac{k\pi}{2n+1})\]
\[=\frac{nC_{2r}^r}{2^{2r}}+\frac{1}{2^{2r-1}}\sum_{j=0}^{r-1}(-1)^{r-j}C_{2r}^j\cdot\frac{1}{2}\left(-1+\frac{\sin((r-j)\pi)}{\sin(\frac{(r-j)\pi}{2n+1})}\right)\]
注意$r-j$为整数，$\sin((r-j)\pi)=0$，故此
\[=\frac{nC_{2r}^r}{2^{2r}}+\frac{1}{2^{2r}}\sum_{j=0}^{r-1}(-1)^{r-j+1}C_{2r}^j\]
\[=\frac{nC_{2r}^r-\frac{C_{2r}^r}{2}}{2^{2r}}\]
\[=\frac{(2n+1)C_{2r}^r}{2^{2r+1}}\]

lemondian

跟这个东东有没有关系？
kuing.cjhb.site/forum.php?mod=viewthread&tid=5332

战巡

这里也同样只示范一个：
注意到
\[\cos((2r-2j)\theta)+i\sin((2r-2j)\theta)=(\cos(\theta)+i\sin(\theta))^{2r-2j}\]
\[2\cos((2r-2j)\theta)=\cos((2r-2j)\theta)+i\sin((2r-2j)\theta)+\cos((2r-2j)\theta)-i\sin((2r-2j)\theta)=(\cos(\theta)+i\sin(\theta))^{2r-2j}+(\cos(\theta)-i\sin(\theta))^{2r-2j}\]
那么
\[2C_{2r}^j\cos((2r-2j)\theta)=C_{2r}^j[(\cos(\theta)+i\sin(\theta))^{2r-2j}+(\cos(\theta)-i\sin(\theta))^{2r-2j}]\]
这里注意
\[\cos(\theta)-i\sin(\theta)=\frac{1}{\cos(\theta)+i\sin(\theta)}\]
于是
\[\mbox{原式}=C_{2r}^j(\cos(\theta)+i\sin(\theta))^{2r-2j}+C_{2r}^{2r-j}(\cos(\theta)+i\sin(\theta))^{2j-2r}\]
\[\sum_{j=0}^{r-1}2C_{2r}^j\cos((2r-2j)\theta)=\sum_{j=0}^{r-1}C_{2r}^j(\cos(\theta)+i\sin(\theta))^{2r-2j}+\sum_{j=0}^{r-1}C_{2r}^{2r-j}(\cos(\theta)+i\sin(\theta))^{2j-2r}\]
\[=\sum_{j=0}^{2r}C_{2r}^j(\cos(\theta)+i\sin(\theta))^{2r-2j}-C_{2r}^r\]
\[=\left(\cos(\theta)+i\sin(\theta)+\frac{1}{\cos(\theta)+i\sin(\theta)}\right)^{2r}-C_{2r}^r\]
\[=(2\cos(\theta))^{2r}-C_{2r}^r\]
即
\[2\sum_{j=0}^{r-1}C_{2r}^j\cos((2r-2j)\theta)=(2\cos(\theta))^{2r}-C_{2r}^r\]
下略