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.

 A: 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.
A: 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$  
