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I could not find the proof of the following formula in the internet:

$$\frac{k\tan x}{\tan kx}=1+\frac{1}{2}\sum_{0<j<k} \frac{\tan x}{\tan (x+\frac{\pi j}{k})}+\frac{\tan x}{\tan (x-\frac{\pi j}{k})}$$

The source I found this formula is here: page 5, proof of (i)

Gabrielek
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user802105
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1 Answers1

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Claim: $$\tan(\theta)+\tan \left(\theta+ \frac{\pi}{n} \right) + \tan\left(\theta + \frac{2\pi}{n}\right) + \dots + \tan \left (\theta + \frac{(n-1)\pi}{n} \right) = -n\cot \left(\frac{n\pi}{2} + n\theta \right)$$

Proof: Sum of tangent functions where arguments are in specific arithmetic series

$$\sum_{0<j<k} {\tan (x+\frac{\pi j}{k})}=-\tan x-k\cot(\frac{k\pi}{2}+kx)$$

$$\sum_{0<j<k} {\tan (x-\frac{\pi j}{k})}=-\sum_{0<j<k} {\tan (-x+\frac{\pi j}{k})}=-\tan x-k\cot(kx-\frac{k\pi}{2})$$

Now remember that $\cot(kx + \frac{k\pi}{2}) = \tan kx $ or -$\tan kx$ depending on wether $k$ is odd or even. So make 2 cases and solve.

  • Thanks a lot for your answer! A related thread: https://math.stackexchange.com/questions/1607134/cotangent-summation-proof – user802105 Jan 05 '21 at 14:49
  • @user802105 You are welcome :). Were you able to solve the problem completely after that? –  Jan 05 '21 at 15:34
  • Yes... There was not much left for me to do, as you had tackled the hardest part. – user802105 Jan 05 '21 at 17:19