0
$\begingroup$

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

  • 1
    See http://math.stackexchange.com/questions/1552737/tan-frac-pi16-tan-frac5-pi16-tan-frac9-pi16-tan-frac13-pi16/1553724#1553724 – 2017-01-06
  • 1
    See also http://math.stackexchange.com/questions/346368/sum-of-tangent-functions-where-arguments-are-in-specific-arithmetic-series – 2017-01-06
  • 0
    thanku lab bhattacharjee but i am not be able to calculate my sum, help me – 2017-01-06
  • 0
    thanks lab bhattacharjee got it. – 2017-01-06

0 Answers 0