Find $\quad\cot^2(2^{\circ})+\cot^2(6^{\circ})+\cot^2({10}^{\circ})++\cdots +\cot^2({86}^{\circ})$

Asked Jul 17 '18 at 20:59

Active Jul 17 '18 at 21:06

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Find $$\cot^2(2^{\circ})+\cot^2(6^{\circ})+\cot^2({10}^{\circ})++\cdots +\cot^2({86}^{\circ})$$

$\mathbf {My Attempt}$
I tried to write the sum backward like this
$$S=\sum_{n=1}^{22} \cot^2(4n-2) = \sum_{n=1}^{22} \cot^2({90}^{\circ}-4n) = \sum_{n=1}^{22} \tan^2(4n)$$ $S=990$ But still can't find any good trigonometry identity to form telescopic series or something like that.
Any hint?

edited Jul 17 '18 at 21:06

asked Jul 17 '18 at 20:59

Wolfdale

See https://math.stackexchange.com/questions/951522/trig-sum-tan-21-circ-tan-22-circ-tan2-89-circ – lab bhattacharjee Jul 18 '18 at 01:35
@lab many thx. I seem I've overlooked this post. I can adapt it now. – Wolfdale Jul 18 '18 at 01:59

Find $\quad\cot^2(2^{\circ})+\cot^2(6^{\circ})+\cot^2({10}^{\circ})++\cdots +\cot^2({86}^{\circ})$

0 Answers0