Finite sum formula for tangent (trigonometry)

[Vahsek]

Hi everyone, I've been looking for the finite sum formulae of trig functions. I've found the easiest ones (sine and cosine). But the one for the tangent seems to be very hard. No mathematical tricks work. Plus I've looked it up on the internet. Nothing. I will greatly appreciate your help. Thanks in advance.

tan x + tan (2x) + tan (3x) + ... + tan (nx) = ?

 

[mathman]

Hi everyone, I've been looking for the finite sum formulae of trig functions. I've found the easiest ones (sine and cosine). But the one for the tangent seems to be very hard. No mathematical tricks work. Plus I've looked it up on the internet. Nothing. I will greatly appreciate your help. Thanks in advance.

tan x + tan (2x) + tan (3x) + ... + tan (nx) = ?

My guess: You are out of luck. I checked Gradshteyn & Ryzhik. They have the sums for sin and cos, as well as sinh and cosh, but nothing for tan.

 

[Vahsek]

My guess: You are out of luck. I checked Gradshteyn & Ryzhik. They have the sums for sin and cos, as well as sinh and cosh, but nothing for tan.

ok, thanks for your consideration though. I'll wait a bit more; maybe someone's got a way to do it.

 

[h6ss]

I found in a textbook that tan(x) can be written as an indefinite sum:

\sum_x \tan ax = i x-\frac1a \psi _{e^{2 i a}}\left(x-\frac{\pi }{2 a}\right) + C \,,\,\,a\ne \frac{n\pi}2 where \psi_q(x) is the q-digamma function.

Computing "sum k from 1 to n of tan(k*x)" in WolframAlpha results into something much more complicated, but an answer is given.

 

[Vahsek]

I found in a textbook that tan(x) can be written as an indefinite sum:

\sum_x \tan ax = i x-\frac1a \psi _{e^{2 i a}}\left(x-\frac{\pi }{2 a}\right) + C \,,\,\,a\ne \frac{n\pi}2 where \psi_q(x) is the q-digamma function.

Computing "sum k from 1 to n of tan(k*x)" in WolframAlpha results into something much more complicated, but an answer is given.

Wow. I had no idea it was that complicated. I'm in high school right now. These functions in real/complex analysis is way beyond me. Anyway, thank you everyone though. At least now I know which direction I must be heading to learn more about it.