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Prove:

$$\cot \frac{\pi}{2m}\cot \frac{2\pi}{2m}\cot \frac{3\pi}{2m}...\cot \frac{(m-1)\pi}{2m}=1$$

This is a roots of unity problem. I managed to show a similar example for $\cos$ by the following:

$z^m-1=0$ with roots $1,e^{2\pi /m},...,e^{2(m-1)\pi /m}$ Put the roots into a polynomial with linear factors:

$(z-1)(z-e^{2\pi /m})...(e^{2(m-1)\pi /m})=z^m-1$ and divide by $z-1$ to get:

$$(z-e^{2\pi /m})...(e^{2(m-1)\pi /m})=1+z+...+z^{m-1}$$

Evaluation at $z=1$ with some complex conjugation (I multiply 2 conjugated equations) yields the result.

I'll be happy to expand on my current work if required.

Edit: Thought I'd add a similar $\sin$ summation

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GRS
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1 Answers1

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We use the fact that $$\left(\cot x\right)\left(\cot\left(\frac{\pi}{2}-x\right)\right)=1.$$ The product of the entry that is $k$ from the beginning and the entry that is $k$ from the end is $1$. (If $m$ is even, there is a "middle" term, but it is $1$.)

André Nicolas
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  • I see, so I simply need to pair up the root according to the formula above? – GRS Apr 28 '16 at 08:39
  • P.S. I changed it to $m$, I typed it up too late yesterday. There are $m-1$ of them, but for the formula to hold I'd have to have even number of $\cot$ functions. This would mean that only odd $m$ will work? – GRS Apr 28 '16 at 08:41
  • Thanks a lot, could you give me some insight into how you came up with the formula? – GRS Apr 28 '16 at 11:14
  • @GRS: Odd number of terms (that is, $m$ even), works too, in that case the middle term is $\cot(\pi/4)$, which is $1$. – André Nicolas Apr 28 '16 at 11:15
  • @GRS: You are welcome. You also know the formula, in the equivalent form $\tan(90^\circ-x)=\frac{1}{\tan x}$. Switch to the complementary angle and the $\tan$ flips. – André Nicolas Apr 28 '16 at 11:21
  • You can also get it from $\sin(\pi/2-x)=\cos x$, $\cos(\pi/2-x)=\sin x$. – André Nicolas Apr 28 '16 at 11:24
  • Thanks, this is much simpler to derive using $\cos$ and $\sin$ – GRS Apr 28 '16 at 13:10
  • @GRS: You are welcome. Note that the basic idea is similar to the one for adding up $1+2+3+\cdots+50$. The sum of the first and last is $51$, as is the sum of second and next to last, and so on.In our case it is multiplication, not addition, but there is a similar "symmetry" thing going on. – André Nicolas Apr 28 '16 at 15:32
  • I tried to derive your results, and it was much easier than expected, almost in a single line :) – GRS Apr 28 '16 at 17:00
  • Yes, this one is very quick. It uses much less machinery that the $\cos$ one referred to in your question. – André Nicolas Apr 28 '16 at 17:05