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How to prove $$ \displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = (2n+1) \prod_{k=1}^{n}\left(1 - \frac{\sin^2 \theta}{\sin^2 \left( \frac{k\pi }{2n+1} \right ) } \right ) $$

So far, I manage to prove $ \displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = (2n+1) \prod_{k=1}^{2n}\left(1 - \frac{\sin \theta}{\sin \left( \frac{k\pi }{2n+1} \right ) } \right ) $ though I am not sure I am aright. enter image description here

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Notice that $sin(-\theta)=-sin(\theta)$, so you have finished the proof! –  awllower Feb 21 '13 at 14:55
    
but $\theta$ goes from $0$ to $\pi$, how to get negative value?? –  hasExams Feb 21 '13 at 14:56
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$sin(\pi-\theta)=sin(\theta)$. Hm. This needs more work to show. Sorry to be too frivolous. If it was $2k\pi /(2n+1)$, then this is fine. Are you sure of the form as stated? –  awllower Feb 21 '13 at 15:01
    
yeah ... it's from this book problem no 1.161 ... I am not sure if it's right though ... –  hasExams Feb 21 '13 at 15:14
    
Did you try using Hadamard factorization theorem? –  Guillermo Feb 21 '13 at 15:35

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up vote 1 down vote accepted

Put $z = e^{i\theta}.$ The LHS becomes $$ \frac{z^{2n+1} - 1/z^{2n+1}}{z-1/z} = \frac{z^{4n+2} - 1}{z^{2n+2} - z^{2n}} = \frac{1}{z^{2n}} \frac{z^{4n+2}-1}{z^2-1}.$$

Let $\zeta_k = e^{\frac{2 \pi i k}{4n+2}} = e^{\frac{\pi i k}{2n+1}} $ be the $k$th root of unity.

The RHS is $$ (2n+1) \prod_{k=1}^n \left( 1 - \frac{\sin^2 \theta}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right) = (2n+1) \prod_{k=1}^n \left(1 - \frac{(z-1/z)^2}{(\zeta_k-1/\zeta_k)^2} \right)\\ = (2n+1) \frac{1}{z^{2n}} \prod_{k=1}^n \left(z^2- \frac{(z^2-1)^2}{(\zeta_k-1/\zeta_k)^2} \right) = (2n+1) \frac{1}{z^{2n}} \prod_{k=1}^n \left(z^2 - \zeta_k^2 \frac{(z^2-1)^2}{(\zeta_k^2-1)^2} \right)$$ So we have an equality between two polynomials that we need to show: $$ \frac{z^{4n+2}-1}{z^2-1} = (2n+1) \prod_{k=1}^n \left(z^2 - \zeta_k^2\frac{(z^2-1)^2}{(\zeta_k^2-1)^2} \right)$$ But these two vanish at the same set of points, namely the roots $\pm\zeta_k$ of unity and their multiplicative inverses $\pm 1/\zeta_k$, where $1\le k\le n$ and are of the same degree ($4n$), so they are scalar multiples of each other. We just need to determine the scalar. To do this, note that $\theta = 0$ was not in fact a singularity of the original LHS since in a neighborhood of zero, we have $$ \frac{\sin (2n+1)\theta}{\sin\theta} \sim 2n+1.$$ The same goes for $\theta = \pi.$

Therefore $z=1$ and $z=-1$ are not singularities of the LHS in $z$ either and we are justified in writing $$ \frac{z^{4n+2}-1}{z^2-1} = z^{4n} + z^{4n-2} + \ldots + z^4 + z^2 + 1.$$ Now the LHS is equal to $2n+1$ at $z=1$, and the product is $$ \left. \prod_{k=1}^n \left(z^2 - \zeta_k^2\frac{(z^2-1)^2}{(\zeta_k^2-1)^2} \right)\right|_{z=1} = 1.$$ Therefore the value of the scalar is $2n+1$ and we are done.

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