1i think after this we need to multipl numerator and denominator with a quantity
such that
the numerator collapses to a single term
just first thoughts
\texttt{deduce that }
\prod_{k=1}^{n-1}{\sin \frac{k\pi}{n}}=\frac{n}{2^{n-1}}
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11 Answers
62hint:
Take z=cos(\pi/n)+isin(\pi/n)
we have to find: \frac{1}{(2i)^n}(z-1/z)(z^2-1/z^2)....(z^n-z^{-n})
After this, there is still some work left to be done.... Which I am not giving away yet... I think it is not very difficult.. but you need to figure out...
162good guess. but i think that will not happen unless n is a prime number... The proof of which, we should not get into right now....
I mean at your level.. If you are interested, read Multiplicative group of integers modulo n
But why i am suggesting you not to read that right now is because you will hardly use that
341\prod_{k=1}^{n-1} \sin \frac{k \pi}{n} = \prod_{k=1}^{n-1} \left(\frac{e^{\frac{ik\pi}{n}} - e^{\frac{ik\pi}{n}}}{2i}\right) = \frac{1}{2^{n-1}} \frac{1}{i^{n-1}} \prod_{k=1}^{n-1} \left(e^{\frac{i2k\pi}{n}}-1\right) \prod_{k=1}^{n-1} \frac{1}{e^{\frac{ik\pi}{n}}}
Now, \prod_{k=1}^{n-1} \frac{1}{e^{\frac{ik\pi}{n}}} = \frac{1}{e^{\frac{i(n-1)\pi}{2}}} = \frac{1}{i^{n-1}}
Also \prod_{k=1}^{n-1} \left( e^{\frac{i2k\pi}{n}}-1 \right) = (-1)^{n-1} \prod_{k=1}^{n-1} \left(1- e^{\frac{i2k\pi}{n}} \right) = (-1)^{n-1} f(1)
where f(z) =1+z+z^2+...+z^{n-1}
and so we have f(1) = n.
Thus the product evaluates to
\frac{1}{2^{n-1}} \ \times \ (-1)^{n-1} \ \times \ \frac{1}{i^{n-1}} \ \times \ \frac{1}{i^{n-1}} \ \times \ n = \frac{1}{2^{n-1}} \ \times \ (-1)^{n-1} \ \times \ \frac{1}{(-1)^{n-1}} \ \times \ n
= \frac{n}{2^{n-1}} as required.
341I remember that akari had asked this problem some weeks ago. This and a similar looking one in the same thread. I had tried to post the solution then, but I was logged out before I could post it, so gave up :D.
62awesome proof..
but a small thing from the rest of you..
Can you see prophet sir's work and complete the rest of the proof.?
1elaboration :
roots of
z^n=1
are
e^{i\frac{2k\pi}{n}},k \in (1,n-1)
(z-1)(1+z+z^2+z^3+.....z^{n-1})=(z-1)(z-\alpha)(z-\alpha^2)........(z-\alpha^{n-1})
putting z=1
n=\prod_{k=1}^{n-1}{(1-e^{i\frac{2k\pi}{n}})}
