# Is there any identity for $\sum_{k=1}^{n}\tan\left(\theta+\frac{k\pi}{\color{red} {2n+1}}\right)$?

Is there any identity for $$\sum_{k=1}^{n}\tan\left(\theta+\frac{k\pi}{\color{red} {2n+1}}\right)$$ or $$\sum_{k=1}^{n}\tan\left(\frac{k\pi}{\color{red} {2n+1}}\right)$$ ?

I thought maybe wrongly that given:

$$\sum_{k=1}^{n}\cot\left(\frac{k\pi}{\color{red} {2n+1}}\right)=\sum_{k=1}^{n}-\tan\left(\frac{\pi}{ {2}}+\frac{k\pi}{\color{red} {2n+1}}\right)$$

and since $$n \cot(nx)$$ is the logarithmic derivative of $$\sin(nx)$$ and $$\cot\left(x+\frac{\pi k}{n}\right)$$ is the logarithmic derivative of $$\sin\left(x+\frac{\pi k}{n}\right)$$, I tried manipulating the identity:

$$2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$$

but I kept getting stuck.

The variation $$\sum_{k=0}^{n-1}\tan\left(\theta+\frac{k\pi}{n}\right)=−n\cot\left(\frac{n\pi}{2}+n\theta\right)$$ is seen to work quite nicely.

• Complex analysis is very useful for those things, caracterizing $\tan(z)$ as the only $\pi$-periodic meromorphic function with simple poles of residue $1$ at $\pi/2+\pi \Bbb{Z}$, analytic and bounded elsewhere and $\lim_{\Im(z)\to +\infty} \tan(z)=i$. Feb 13, 2022 at 15:37
• @reuns thank you I've updated the tag to complex-analysis and will try to give some thought on that idea you mention. Feb 13, 2022 at 15:55
• The reason why you have an identity for sine is that sine function can be easily rewrite as a simple sum of power function. Feb 13, 2022 at 18:14
• Mathematica says for general a and b $$\sum _{k=1}^n \tan (a k+b)=i n+\frac{2 i \psi _{e^{2 i a b}}\left(1-\frac{i \pi }{\ln \left(e^{2 i a b}\right)}\right)}{\ln \left(e^{2 i a b}\right)}-\frac{2 i \psi _{e^{2 i a b}}\left(1+n-\frac{i \pi }{\ln \left(e^{2 i a b}\right)}\right)}{\ln \left(e^{2 i a b}\right)}$$ where: $\psi _{e^{2 i a b}}\left(1+n-\frac{i \pi }{\ln \left(e^{2 i a b}\right)}\right)$ is QPolyGamma function. Feb 15, 2022 at 17:13
• @High GPA on the balance of things this does seem to be true. Feb 21, 2022 at 18:07

Disclaimer: I thought this would turn out to be a routine, olympiad type exercise on manipulating roots of unity and so I started writing an answer but then I got stuck. Leaving it here for now if someone else/myself manage to fix it and will remove it before the bounty period ends.

Let $$\theta = \dfrac{\pi}{2n+1}.$$ Then,

$$i\tan\theta = \dfrac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = 1 - \dfrac{1}{1+e^{2i\theta}},$$ so it suffices to find them sum: $$\sum_{k=1}^n\dfrac{1}{1+e^{2ik\theta}}.$$ If we let $$z_k = e^{2ik\theta}$$, then the plan of attack is we find what unique, monic, degree $$n$$ polynomial have $$z_1,z_2,\dots z_n$$ as roots. If we could find a simple expression $$f(z),$$ then $$f(z-1)$$ has roots $$1+z_k$$'s and from there we can write a simple Vieta ratio to find the desired sum.

Note that $$z_k = e^{2\pi i\tfrac{k}{2n+1}} = \rho^{k},$$ where is a primitive root of unity of order $$2n+1.$$ This means that:

$$\dfrac{z^{2n+1}-1}{z-1} = \prod_{k=1}^n(z-\rho^k)\prod_{k=n+1}^{2n}(z-\rho^k)=f(z) \prod_{k=1}^n(z-\rho^{k+n}).$$ Some basic manipulation tells us: $$\prod_{k=1}^n(z-\rho^{k+n}) = \prod_{k=1}^n(z-\rho^{2n+1-k})=\prod_{k=1}^n(z-\rho^{-k}) = \rho^{-n(n+1)/2}\prod_{k=1}^n\rho^{k}(z-\rho^{-k}) =$$ $$=\rho^{-n(n+1)/2}\cdot (-1)^n\prod_{k=1}^n(1-\rho^kz) = \rho^{-n(n+1)/2}\cdot (-z)^n\prod_{k=1}^n\left(\frac 1z-\rho^k\right) =$$ $$\rho^{-n(n+1)/2}\cdot (-z)^n f\left(\frac 1z\right),$$ so we obtain: $$z^nf(z)f\left(\frac{1}{z}\right) = \dfrac{z^{2n+1}-1}{z-1}\cdot\rho^{n(n+1)/2}\cdot (-1)^n.$$

• Thank you for your effort sketching this out it helps me understand the problem much clearer than before and the difficulties. I have since yesterday found that $\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)=n(2n+1)$ works as shown here math.stackexchange.com/questions/173447/… and here Feb 14, 2022 at 8:57
• math.stackexchange.com/questions/2339/… but as robjohn and yourself has explained it is not so simple in this case. My attempts were in manipulating the sine product into sum tangent but I could only make progress with the sine squared. Feb 14, 2022 at 8:57
• @onepound yes I quickly rediscovered that after writing this. At this point, my feeling is it does not admit a simple formula. I also considered the Chebyshev polynomial of second kind $$2^{2n}U_{2n}(x) = \prod _{k=0}^{2n}\left(x-\cos\dfrac{k\pi}{2n+1}\right),$$ which you can manipulate further to get $\tan\dfrac{k\pi}{2n+1}$ on the right hand side, but that only yielded similar equation as the one above I wrote. Feb 14, 2022 at 18:05
• How many olympiad problems are about calculating sums using complex numbers? Feb 19, 2022 at 13:33
• @AitorIribarLopez I definitely learned Chebyshev and Cyclotomic polynomials as part of IMO training. I guess most of the problems won't be just about summing things as that would be too simple at IMO level. But cyclotomic polynomials solve a lot of number theory problems. There was a problem about summing some inverse cosine's in my TST one time which I failed big time because I had not seen Chebyshev in 9th grade. Feb 19, 2022 at 19:52