This equality in the title is one answer in the MSE post Funny Identities. At first, I thought it had to do with $7$ being a Mersenne prime, but a little experimentation with Mathematica's integer relations found,

$$\frac{1}{\sin(2\pi/15)} + \frac{1}{\sin(4\pi/15)} + \frac{1}{\sin(7\pi/15)} = \frac{1}{\sin(\pi/15)}$$

$$\frac{1}{\sin(2\pi/31)} + \frac{1}{\sin(4\pi/31)} + \frac{1}{\sin(8\pi/31)} + \frac{1}{\sin(15\pi/31)} = \frac{1}{\sin(\pi/31)}$$

so the Mersenne number need not be prime. Let $M_n = 2^n-1$. How do we prove that,

$$\frac{1}{\sin(M_{n-1}\pi/M_n)}+\sum_{k=1}^{n-2} \frac{1}{\sin(2^k\pi/M_n)} = \frac{1}{\sin(\pi/M_n)}$$

indeed holds true for all integer $n>2$?

Edit (an hour later):

I just realized that since, for example, $\sin(3\pi/7)=\sin(4\pi/7)$, then the question can be much simplified as,

$$\sum_{k=1}^{n-1} \frac{1}{\sin(2^k\pi/M_n)} \overset{?}{=} \frac{1}{\sin(\pi/M_n)}$$

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    $\begingroup$ Oops, thanks for the formatting correx, Micah. $\endgroup$ – Tito Piezas III Sep 28 '13 at 1:21

Let $$\displaystyle S=\sum_{k=1}^{n-1}\csc \left(\frac{2^k \pi}{2^n-1} \right)$$ Using Euler's Formula, we can express this sum in terms of complex numbers. $$S=\sum_{k=1}^{n-1} \frac{2i}{e^{i 2^k \pi/(2^n-1)}-e^{-i2^k\pi /(2^n-1)}}=2i \sum_{k=1}^{n-1}\frac{e^{i 2^k \pi/(2^n-1)}}{e^{i 2^{k+1}\pi/(2^n-1)}-1}$$ For simplicity, let us assume $x=e^{i\pi/(2^n-1)}$. Then $$\begin{align*} S &= 2i\sum_{k=1}^{n-1}\frac{x^{2^k}}{x^{2^{k+1}}-1} \\ &= 2i \sum_{k=1}^{n-1}\frac{(x^{2^k}+1)-1}{(x^{2^k}+1)(x^{2^k}-1)} \\ &= 2i \sum_{k=1}^{n-1}\left( \frac{1}{x^{2^k}-1}-\frac{1}{x^{2^{k+1}}-1}\right) \end{align*}$$ This is a telescoping sum and it's value is $$\begin{align*} S &= 2i \left( \frac{1}{x^2-1}-\frac{1}{x^{2^n}-1}\right) \end{align*}$$ Back substituting, gives $$\begin{align*} S &= 2i \left( \frac{1}{e^{2i\pi/(2^n-1)}-1}-\frac{1}{e^{i 2^n \pi/(2^n-1)}-1}\right) \\ &= 2i \left( \frac{1}{e^{2i\pi/(2^n-1)}-1}+\frac{1}{e^{i\pi/(2^n-1)}+1}\right)\\ &= 2i \frac{e^{i\pi/(2^n-1)}}{e^{2i\pi/(2^n-1)}-1} \\ &= \csc \left( \frac{\pi}{2^n-1}\right) \end{align*}$$ as desired.

  • $\begingroup$ dang, you got it just before I did, though I think mine's more pleasing :) $\endgroup$ – Calvin Lin Sep 28 '13 at 3:45
  • $\begingroup$ @Integral: I've accepted it since you answered first, and it is quite detailed. (Btw, you really should use your real name in this forum. This place is moderated anyway, so people are relatively civil.) $\endgroup$ – Tito Piezas III Sep 28 '13 at 15:08

I would suggest a further simplification of your problem, namely that $ \frac{1}{ \sin (\pi/ M_n) } = - \frac{1}{ \sin ( 2^n \pi / M_n)} $. The identity then becomes

$$ \sum_{i=1}^n \frac{1}{\sin (2^n \pi / M_n )} = 0. $$

We now use

$$ \frac{1}{\sin \theta} = \cot \frac{\theta}{2} - \cot \theta,$$

which you can verify for yourself, to conclude that the result follows from telescoping, since $ \cot \frac{ \pi}{2^n - 1} = \cot \frac{ 2^n \pi } { 2^n - 1}$.

Note: I got the trig identity from Wikipedia trig identity, knowing that I wanted $\csc \theta$ and something that could telescope.


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