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Prove that $$\frac{1}{\sin^{2}\frac{\pi }{4k+2}}+\frac{1}{\sin^{2}\frac{3\pi }{4k+2}}+\frac{1}{\sin^{2}\frac{5\pi }{4k+2}}+\cdots+\frac{1}{\sin^{2}\frac{(2k-1)\pi }{4k+2}}=2k(k+1)$$

Hypergeometricx
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mja
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    Closely related question: http://math.stackexchange.com/questions/544228/is-sum-k-1m-1-frac1-sin2-frack-pim-fracm2-13-true-for-m?lq=1. It may help. – Oleg567 Aug 13 '15 at 17:42
  • Using $$2\sin^2A=1-\cos2A$$ related to http://math.stackexchange.com/questions/1351337/product-of-cosines-prod-r-17-cos-fracr-pi15/1352365#1352365 – lab bhattacharjee Aug 13 '15 at 17:54
  • It appears that the $k$s in the expanded sum should be $n$s. – robjohn Aug 14 '15 at 16:45
  • Agree with @robjohn. Or, in the question header, the $k$'s should be $j$'s, and the $n$'s should be $k$'s. – Hypergeometricx Aug 14 '15 at 17:01

1 Answers1

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Using $(7)$ from this answer, we get $$ \begin{align} \sum_{k=1}^n\frac1{\sin^2\left(\frac{2k-1}{4n+2}\pi\right)} &=\sum_{k=1}^n\csc^2\left(\frac{2k-1}{2n+1}\frac\pi2\right)\tag{1}\\ &=\sum_{k=1}^n\sec^2\left(\frac{2n-2k+2}{2n+1}\frac\pi2\right)\tag{2}\\ &=\sum_{k=1}^n\sec^2\left(\frac{k}{2n+1}\pi\right)\tag{3}\\ &=n+\sum_{k=1}^n\tan^2\left(\frac{k}{2n+1}\pi\right)\tag{4}\\[4pt] &=n+n(2n+1)\tag{5}\\[12pt] &=2n(n+1)\tag{6} \end{align} $$ Explanation:
$(1)$: $\frac1{\sin(x)}=\csc(x)$
$(2)$: $\csc(x)=\sec\left(\frac\pi2-x\right)$
$(3)$: substitute $k\mapsto n+1-k$
$(4)$: $\sec^2(x)=1+\tan^2(x)$
$(5)$: use $(7)$ from this answer
$(6)$: simplify

Community
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robjohn
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  • Would the downvoter care to comment? – robjohn Aug 15 '15 at 01:43
  • Would the downvoter care to comment? they never do that's problem – onepound Feb 13 '22 at 16:43
  • It's been over $6$ years, so I don't think they will. The main motivation for someone to leave a critical comment would be so that the post could be improved. Of course, if someone leaves a critical comment, it is not anonymous, and they may fear reprisal. A glance at my profile would show there is little chance of retribution. – robjohn Feb 13 '22 at 18:57
  • I tool an interest in your excellent answer as I hoped it would solved my question https://math.stackexchange.com/questions/4379583/is-there-any-identity-for-sum-k-1n-tan-left-theta-frack-pi-colorred only I'm new to contour integration so not sure if it is applicable to $\sum_{k=1}^{n}\tan\left(\theta+\frac{k\pi}{\color{red} {2n+1}}\right)$ – onepound Feb 13 '22 at 19:03
  • This method gives the sum of the squares of the tangents, but the sum of the tangents is not so easy. – robjohn Feb 13 '22 at 21:40
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    The reason we can sum the squares of the tangents and not the tangents this way, is that $\tan(-x)=-\tan(x)$ so the sum of the tangents around the circle cancels to $0$; whereas $\tan^2(-x)=\tan^2(x)$ so the sum of the squares of the tangents around the circle don't cancel and we can sum around the whole circle and divide by $2$. – robjohn Feb 13 '22 at 21:48
  • Thank you for explaining the method and why it works for squares of tangents. Nevertheless even if the method is not applicable to my question it is still interesting. – onepound Feb 14 '22 at 08:44