# Why is: \sum_{k=1}^{[\frac{n}{2}]}\sin^2((2k-1)\frac{\pi}{n})=\frac{n}{4}

Why is:
$\sum _{k=1}^{\left[\frac{n}{2}\right]}{\mathrm{sin}}^{2}\left(\left(2k-1\right)\frac{\pi }{n}\right)=\frac{n}{4}$
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chezpepina87j
Let
$T=\sum _{k=1}^{\left[\frac{n}{2}\right]}\mathrm{cos}\left(\left(2k-1\right)\frac{2\pi }{n}\right)$
$S=\sum _{k=1}^{\left[\frac{n}{2}\right]}{\mathrm{sin}}^{2}\left(\left(2k-1\right)\frac{\pi }{n}\right)$
We can get the relation $T+2S=\left[\frac{n}{2}\right]$
If n is even then using summation of cosines whose angles are in AP we get,
$T=\sum _{k=1}^{\left[\frac{n}{2}\right]}\mathrm{cos}\left(\left(2k-1\right)\frac{2\pi }{n}\right)=\frac{\mathrm{sin}\left[\frac{n}{2}\right]\frac{2\pi }{n}}{\mathrm{sin}\frac{2\pi }{n}}\mathrm{cos}\left[\frac{2\pi }{n}+\left(\left[\frac{n}{2}\right]-1\right)\frac{2\pi }{n}\right]=0$
$⇒S=\frac{n}{4}$
If n is odd then let $n=2m+1$. We get,
$T=\sum _{k=1}^{m}\mathrm{cos}\left(\left(2k-1\right)\frac{2\pi }{2m+1}\right)$
$=\frac{\mathrm{sin}\frac{2\pi m}{2m+1}}{\mathrm{sin}\frac{2\pi }{2m+1}}\mathrm{cos}\left[\frac{2\pi }{2m+1}+\left(m-1\right)\frac{2\pi }{2m+1}\right]$
$=\frac{\mathrm{sin}\frac{2\pi m}{2m+1}\mathrm{cos}\frac{2\pi m}{2m+1}}{\mathrm{sin}\frac{2\pi }{2m+1}}$
$=\frac{-\mathrm{sin}\frac{2\pi }{2m+1}}{2\mathrm{sin}\frac{2\pi }{2m+1}}=\frac{-1}{2}$
$⇒S=\frac{2m+1}{2}=\frac{n}{4}$