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

Question

A problem I solved previously I get

$$\frac{\sin(n\theta)}{\sin(\theta)} = 2^{n-1}\prod_{k=1}^{n-1}\bigg(\cos(\theta)-\cos{\bigg(\frac{k\pi}{n}}\bigg)\bigg)$$

So considering $z^{4n+2} = 1$ leads to

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

Now

$$\cos\bigg(\frac{2n\pi}{2n+1}\bigg) = \cos\bigg(\pi-\frac{\pi}{2n+1}\bigg) =-\cos\bigg(\frac{\pi}{2n+1}\bigg)$$

$$\cos\bigg(\frac{(2n-1)\pi}{2n+1}\bigg) = \cos\bigg(\pi-\frac{2\pi}{2n+1}\bigg) =-\cos\bigg(\frac{2\pi}{2n+1}\bigg)$$

$$...........................................................$$

$$\cos\bigg(\frac{(n+1)\pi}{2n+1}\bigg) = \cos\bigg(\pi-\frac{n\pi}{2n+1}\bigg) =-\cos\bigg(\frac{n\pi}{2n+1}\bigg)$$

multiplying terms $k = 1$ with $k = 2n, k =2$ with $k = 2n-1\dots k = n$ with $k = n+1$ terms, from$......(1)$ I get

$$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = 2^{2n}\cdot\bigg(\cos(\theta)+\cos\bigg(\frac{\pi}{2n+1}\bigg)\bigg)\cdot \bigg(\cos(\theta)-\cos\bigg(\frac{\pi}{2n+1}\bigg)\bigg)\hspace{82pt}\cdot\bigg(\cos(\theta)+\cos\bigg(\frac{2\pi}{2n+1}\bigg)\bigg)\cdot\bigg(\cos(\theta)-\cos\bigg(\frac{2\pi}{2n+1}\bigg)\bigg)........\hspace{56pt}\bigg(\cos(\theta)+\cos\bigg(\frac{n\pi}{2n+1}\bigg)\bigg)\cdot\bigg(\cos(\theta)-\cos\bigg(\frac{n\pi}{2n+1}\bigg)\bigg)$$

$$ \hspace{35pt}= 2^{2n}\prod_{k = 1}^{n} \bigg(\cos^{2}(\theta)-\cos^{2}\bigg(\frac{k\pi}{2n+1}\bigg)\bigg).............(2)$$

How can I prove $$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = (2n+1)\prod_{k = 1}^{n}\bigg(1-\frac{\sin^2(\theta)}{\sin^2\big(\frac{k\pi}{2n+1}\big)}\bigg)$$ from my calculation? Is my way of approaching this problem is ok or not? If it isn't give me one or a few hints.

Answer 1

I figured how to do it

from $.....(2)$ we get-

$$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = 2^{2n}\prod_{k=1}^{n}\bigg(1-sin^{2}(\theta)-1+sin^2 \bigg(\frac{k\pi}{2n+1}\bigg)\bigg)$$

$$\hspace{60pt} = 2^{2n}\prod_{k=1}^{n}\sin^{2}\bigg(\frac{k\pi}{2n+1}\bigg)\bigg(1-\frac{\sin^2(\theta)}{\sin^{2}\big(\frac{k\pi}{2n+1})}\bigg)$$

$$\hspace{79pt}=2^{2n}\prod_{k=1}^{n}\sin^{2}\bigg(\frac{k\pi}{2n+1}\bigg)\prod_{k=1}^{n}\bigg(1-\frac{\sin^2(\theta)}{\sin^{2}\big(\frac{k\pi}{2n+1})}\bigg)$$ $$\hspace{320pt}...........(3)$$

Now from here we know $$\prod_{k=1}^{n-1}\sin \bigg(\frac{k\pi}{n}\bigg) = \frac{n}{2^{n-1}}$$

putting $2n+1$ instead of $n$ we get $$\hspace{5pt}\prod_{k=1}^{2n}\sin \bigg(\frac{k\pi}{2n+1}\bigg) = \frac{2n+1}{2^{2n}}$$

$$\Rightarrow2^{2n}\prod_{k=1}^{2n}\sin \bigg(\frac{k\pi}{2n+1}\bigg) = 2n+1$$ $$\hspace{260pt}..............(4)$$

$$\sin\bigg(\frac{2n\pi}{2n+1}\bigg) = \sin\bigg(\frac{(2n+1)\pi-\pi}{2n+1}\bigg) = \sin\bigg(\pi-\frac{\pi}{2n+1}\bigg) = \sin\bigg(\frac{\pi}{2n+1}\bigg)$$

$$\sin\bigg(\frac{(2n-1)\pi}{2n+1}\bigg) = \sin\bigg(\frac{(2n+1)\pi-2\pi}{2n+1}\bigg) = \sin\bigg(\pi-\frac{2\pi}{2n+1}\bigg) = \sin\bigg(\frac{2\pi}{2n+1}\bigg)$$ $$.............................................................................$$

$$\sin\bigg(\frac{(n+1)\pi}{2n+1}\bigg) = \sin\bigg(\frac{(2n+1)\pi-n\pi}{2n+1}\bigg) = \sin\bigg(\pi-\frac{n\pi}{2n+1}\bigg) = \sin\bigg(\frac{n\pi}{2n+1}\bigg)$$

So from...(4) we get

$$2^{2n}\prod_{k=1}^{n}\sin^{2}\bigg(\frac{k\pi}{2n+1}\bigg) = 2n+1$$

and

$$\frac{\sin((2n+1)\theta)}{\sin(\theta)} = (2n+1)\prod_{k=1}^{n}\bigg(1-\frac{\sin^2(\theta)}{\sin^{2}\big(\frac{k\pi}{2n+1})}\bigg)$$ $$[proved]$$