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$$\sum_{n=1}^{11}\sin^{14}\left(\theta+\frac{2n\pi}{11}\right)=?$$ By wolfram alpha, we know that $$sin^{14}x=\frac{1}{8192}(-3003\cos(2x)+2002\cos(4x)-1001\cos(6x)+364\cos(8x)-91\cos(10x)+14\cos(12x)-\cos(14x)+1716).$$ And by the formula : $$\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+n\beta)=\frac{\sin\left(\frac{(n+1)\beta}{2}\right)\sin\left(\alpha+\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}$$ , we can find the closed form of above series in terms of $\theta$. But this method is very slow. Have another faster ways ? I think it maybe use "roofs of polynomial" technic. Thank you.

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Have you tried using the fact that $\left(e^{i\left(\theta + \frac{2n\pi}{11}\right)}\right)^{14} = 1$ for all $n = 0, 1,2,...,11$? –  Lee Yiyuan Apr 19 at 8:10
    
According to wolfram alpha, $\sum_{n=1}^{11}\cos(\theta + \frac{2n\pi}{11}) = 0$ (as well as whenever the inner term of the cosine is multiplied by any integer), which I suppose would mean that this sum would actually just equal $11 * 1716/8192$? It sounds weird to me but wolfram alpha usually knows what its talking about. –  KSab Apr 19 at 8:12
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Sum of 11th roots of unity is 0. –  Awesome Apr 19 at 10:06

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Put $e^{2\pi i/11}=:\omega$ and $e^{i\theta}\omega^n=:z_n$. Then $$\sin\left(\theta+{2n\pi\over 11}\right)={\rm Im}(z_n)={1\over 2i}\bigl(e^{i\theta}\omega^n-e^{-i\theta}\omega^{-n}\bigr)$$ and $$\sin^{14}\left(\theta+{2n\pi\over 11}\right)={-1\over 2^{14}}\sum_{k=0}^{14}(-1)^k{14\choose k}e^{i(14-2k)\theta}\>\omega^{(14-2k)n}\ .$$ Now $$\sum_{n=1}^{11}\omega^{(14-2k)n}=0\ ,$$ unless $14-2k$ is divisible by $11$. The latter happens iff $k=7$, in which case $\sum_{n=1}^{11}\omega^{(14-2k)n}=11$. It follows that $$\sum_{n=1}^{11}\sin^{14}\left(\theta+{2n\pi\over 11}\right)={11\over 2^{14}}{14\choose 7}={4719\over2048}\ .$$

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thank you very much –  kong Apr 19 at 10:56

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