Trigonometric sum of tangents series

Asked Jan 06 '17 at 08:16

Active Jan 06 '17 at 08:53

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Trigonometric sum of series $\displaystyle \tan \left(\frac{\pi}{15}\right)+\tan \left(\frac{4\pi}{15}\right)+\tan \left(\frac{7\pi}{15}\right)+\tan \left(\frac{10\pi}{15}\right)+\tan \left(\frac{13\pi}{15}\right)$

Substituting $\displaystyle \frac{\pi}{15} = \theta$, then $\displaystyle \pi=15 \theta$ so $7\theta = \pi-8 \theta$

could some help me with this??, thanks

edited Jan 06 '17 at 08:53

asked Jan 06 '17 at 08:16

DXT

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See http://math.stackexchange.com/questions/1552737/tan-frac-pi16-tan-frac5-pi16-tan-frac9-pi16-tan-frac13-pi16/1553724#1553724 – lab bhattacharjee Jan 06 '17 at 08:37
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See also http://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series – Alex Silva Jan 06 '17 at 08:57
thanku lab bhattacharjee but i am not be able to calculate my sum, help me – DXT Jan 06 '17 at 09:56
thanks lab bhattacharjee got it. – DXT Jan 06 '17 at 13:58

Trigonometric sum of tangents series

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