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The question is to evaluate $$\frac{\sum_{k=0}^{6}\csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}$$ where $a=\pi/8$ without looking at the trigonometric table.

I tried to transform the $\csc^2$ term to $\cot^2$ term and use addition formula but it made the problem too cumbersome.I also tried to manipulate the numerator in the form so that it telescopes but couldnot succeed.I am not in need of a full solution.Any idea or hint to proceed shall be highly appreciated.Thanks.

Reese Johnston
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Navin
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  • It might help to know that the sum has the same value for all $a$ for which the denominator is not $0$. The general result is $\frac{\sum_{k=0}^{n-1}\csc^2(x + \frac{k\pi}{n})}{n^2\csc^2(n x)} = 1$ valid for all $n$. This question might be useful: http://math.stackexchange.com/questions/217240/reference-for-a-tangent-squared-sum-identity See also http://math.stackexchange.com/questions/1657093/reference-for-finit-sum-of-cotangents – Winther Feb 13 '17 at 01:51
  • See also : http://math.stackexchange.com/questions/1562037/question-regarding-fn-cot2-left-frac-pi-n-right-cot2-left-frac2-pin?noredirect=1&lq=1 http://math.stackexchange.com/questions/45144/proving-frac1-sin2-frac-pi14-frac1-sin2-frac3-pi14 – lab bhattacharjee Feb 13 '17 at 05:40
  • @labbhattacharjee thanks for the links.They were very much helpful . – Navin Feb 13 '17 at 11:50

1 Answers1

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The result of the summation is:

$\frac{\sum_{k=0}^{6}\csc^2(a+\frac{k\pi}{7})}{7\csc^2(7a)}=7$.

The summation, after some calculation, is reduced to:

$\frac{(A E F+B D F+C D E)+D E F}{7 D E F}=7$

where

$A=4-2\sqrt{2}+(2\sqrt{2}-2)cos(\frac{\pi}{7})$,

$B=-4+2\sqrt{2}+(2-2\sqrt{2})sin(\frac{\pi}{14})$,

$C=-4+2\sqrt{2}+(-2+2\sqrt{2})sin(\frac{3\pi}{14})$,

$D=2\sqrt{2} cos(\frac{\pi}{7})+ sin(\frac{3\pi}{14})+2$,

$E= cos(\frac{\pi}{7})- 2\sqrt{2} sin(\frac{\pi}{14})-2$,

$F=\sqrt{2} sin(\frac{3\pi}{14})+ sin(\frac{\pi}{14})-2$.

By substituting and grouping we get:

$\frac{6+\frac{1}{8}}{7 \frac{1}{8}}=7$.

giuseppe mancò
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