how to prove $\displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = …$

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.

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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
$sin(\pi-\theta)=sin(\theta)$. 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

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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First consider the following Lemma

Lemma :

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

Proof : Note that,

$\displaystyle \prod_{k=1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right) = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=n+1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right)$

$\displaystyle = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=1}^{n} \sin \left(\dfrac{(n+k)\pi}{2n+1}\right)$

$\displaystyle = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \ \left(\because \prod_{k=1}^{n} f(k) = \prod_{k=1}^{n} f(n+1-k) \right)$

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

But,

$\displaystyle \prod_{k=1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}}$ (For my proof of this, see here )

$\displaystyle \implies \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}}$

Now, let $\displaystyle \text{P} = (2n+1)\sin(\theta) \prod_{k=1}^n \left(1 -\dfrac{\sin^2(\theta)}{\sin^2\left(\frac{k\pi}{2n+1}\right)} \right)$

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

$\displaystyle = 2^{2n} \sin(\theta) \prod_{k=1}^n \left( \cos^2(\theta) - \cos^2\left(\frac{k\pi}{2n+1}\right) \right)$ (Using the Lemma)

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

$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) - \cos \left(\frac{(2n+1 - k)\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) \ \left(\because \cos (\pi -x) = -\cos x \right)$

$\displaystyle = 2^{2n} \sin(\theta) \left(\prod_{k=1}^n \left( \cos (\theta) - \cos \left(\frac{(n + k)\pi}{2n+1}\right) \right)\right) \cdot \left( \prod_{k=1}^n \left( \cos(\theta) - \cos \left(\frac{k\pi}{2n+1}\right) \right) \right) \ \left(\because \prod_{k=1}^{n} f(k) = \prod_{k=1}^{n} f(n+1-k) \right)$

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

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

Also,

$\displaystyle U_{n} (x) = 2^{n} \prod_{k=1}^{n} \left(x - \cos \left(\frac{k\pi}{n+1}\right) \right)$

where $\displaystyle U_{n} (x)$ denotes the Chebyshev Polynomial of the Second kind.

$\displaystyle \implies \text{P} = 2^{2n} \sin(\theta) \cdot 2^{-2n} \cdot U_{2n} (\cos \theta)$

$\displaystyle = \sin ((2n+1) \theta) \ \left(\because U_{n} (\cos \theta) = \dfrac{\sin ((n+1) \theta)}{\sin \theta} \right) \quad \square$

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