Prove that $\mathrm{cos}\frac{2\pi}{2n+1}+\mathrm{cos}\frac{4\pi}{2n+1}+\mathrm{cos}\frac{6\pi}{2n+1}+...+\mathrm{cos}\frac{2n\pi}{2n+1}=\frac{-1}{2}$

rjawbreakerca
2022-07-03
Answered

Prove that $\mathrm{cos}\frac{2\pi}{2n+1}+\mathrm{cos}\frac{4\pi}{2n+1}+\mathrm{cos}\frac{6\pi}{2n+1}+...+\mathrm{cos}\frac{2n\pi}{2n+1}=\frac{-1}{2}$

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In any triangle is $\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C=\frac{3\sqrt{3}}{2}$ always
Well, I came with an interesting proof. But I just want to verify it
Applied at function $y=\mathrm{sin}x$
We have, $\mathrm{sin}(\frac{A+B+C}{2})\ge \frac{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}{3}$
From here we will get
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Also by A.M. $\ge $ G.M. in an acute angled triangle
$\frac{\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C}{3}\ge \sqrt[3]{\mathrm{sin}A\mathrm{sin}B\mathrm{sin}C}$
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and from this I get
$\mathrm{sin}A+\mathrm{sin}B+\mathrm{sin}C\ge \frac{3\sqrt{3}}{2}$

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