Alternate sum with sins
Alternate sum with sins
Show that \(\displaystyle \sum_{k=1}^{n-1}(-1)^k\sin^n\left(\frac{k\pi}{n}\right)=\frac{(1+(-1)^n)n}{2^n}\cos\left(\frac{n\pi}{2}\right).\)
(Ovidiu Furdui)
(Ovidiu Furdui)
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