Question

Suppose $\sin(\theta)≠0$. Prove that
$\frac{ \sin(n \theta) }{ \sin( \theta) } = {2}^{n - 1} \prod \limits ^{n - 1}_{k = 1} \{ \cos( \theta) - \cos( \frac{k\pi}{n} ) \}$
​

Answer 1

We know,

$\small\longrightarrow e^{i\theta}=\cos\theta+i\sin\theta$

And,

$\small\longrightarrow e^{-i\theta}=\cos\theta-i\sin\theta$

Adding both,

$\small\text{ \longrightarrow\cos\theta=\dfrac{e^{i\theta}+e^{-i\theta}}{2} }$

Taking difference,

$\small\text{ \longrightarrow\sin\theta=\dfrac{e^{i\theta}-e^{-i\theta}}{2i} }$

So,

$\small\text{ \longrightarrow\dfrac{\sin(n\theta)}{\sin\theta}=\dfrac{\left(\dfrac{e^{in\theta}-e^{-in\theta}}{2i}\right)}{\left(\dfrac{e^{i\theta}-e^{-i\theta}}{2i}\right)} }$

$\small\text{ \longrightarrow\dfrac{\sin(n\theta)}{\sin\theta}=\dfrac{e^{in\theta}-\dfrac{1}{e^{in\theta}}}{e^{i\theta}-\dfrac{1}{e^{i\theta}}} }$

$\small\text{ \longrightarrow\dfrac{\sin(n\theta)}{\sin\theta}=\dfrac{\left(\dfrac{e^{2in\theta}-1}{e^{in\theta}}\right)}{\left(\dfrac{e^{2i\theta}-1}{e^{i\theta}}\right)} }$

$\small\text{ \longrightarrow\dfrac{\sin(n\theta)}{\sin\theta}=\dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}\cdot e^{i(1-n)\theta}\quad\quad\dots(1) }$

We know,

$\small\text{ \longrightarrow e^{i2\pi}=\left(e^{\frac {i2\pi}{n}}\right)^n=1 }$

$\small\text{ \longrightarrow \left(e^{\frac {i2\pi k}{n}}\right)^n=1^k=1 }$

Taking $\small\text{ x=e^{\frac{i2\pi k}{n}}, }$

$\small\longrightarrow x^n-1=0$

This means $\small\text{ e^{\frac{i2\pi k}{n}} }$ is the k'th root of $\smallx^n-1=0$ so $\smallk=an+r$ where $\smalla,\ r\in\mathbb{Z}$ and $\small0\leq r<n.$

So one can write,

$\small\text{ \displaystyle\longrightarrow x^n-1=\prod_{k=0}^{n-1}\left(x-e^{\frac{i2\pi k}{n}}\right) }$

Taking $\smallx=e^{2i\theta},$

$\small\text{ \displaystyle\longrightarrow\left(e^{2i\theta}\right)^n-1=\prod_{k=0}^{n-1}\left(e^{2i\theta}-e^{\frac{i2\pi k}{n}}\right) }$

$\small\text{ \displaystyle\longrightarrow e^{2in\theta}-1=\left(e^{2i\theta}-1\right)\prod_{k=1}^{n-1}\left(e^{2i\theta}-e^{\frac{i2\pi k}{n}}\right) }$

$\small\text{ \displaystyle\longrightarrow\dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}=\prod_{k=1}^{n-1}\left(e^{i\theta}+e^{\frac{i\pi k}{n}}\right)\left(e^{i\theta}-e^{\frac{i\pi k}{n}}\right) }$

$\small\text{ \displaystyle\longrightarrow\dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}=\prod_{k=1}^{n-1}\left(e^{i\theta}-\left(-e^{\frac{i\pi k}{n}}\right)\right)\left(e^{i\theta}-e^{\frac{i\pi k}{n}}\right) }$

$\small\text{ \displaystyle\longrightarrow\dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}=\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{-i\pi}\cdot e^{\frac{i\pi k}{n}}\right)\left(e^{i\theta}-e^{\frac{i\pi k}{n}}\right) }$

$\small\text{ \displaystyle\longrightarrow\dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}=\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{-\frac{i(n-k)\pi}{n}}\right)\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{\frac{i\pi k}{n}}\right) }$

The terms in $\small\text{ \displaystyle\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{-\frac{i(n-k)\pi}{n}}\right) }$ can be reversed and become $\small\text{ \displaystyle\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{-\frac{ik\pi}{n}}\right). }$

$\small\text{ \displaystyle\longrightarrow \dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}=\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{-\frac{ik\pi}{n}}\right)\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{\frac{ik\pi}{n}}\right) }$

$\small\text{ \displaystyle\longrightarrow \dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}=\prod_{k=1}^{n-1}\left(e^{2i\theta}-e^{i\theta}\left(e^{\frac{ik\pi}{n}}+e^{-\frac{ik\pi}{n}}\right)+1\right) }$

$\small\text{ \displaystyle\longrightarrow \dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}=\prod_{k=1}^{n-1}e^{i\theta}\left(e^{i\theta}-e^{\frac{ik\pi}{n}}-e^{-\frac{ik\pi}{n}}+e^{-i\theta}\right) }$

$\small\text{ \displaystyle\longrightarrow \dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}=e^{i(n-1)\theta}\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{\frac{ik\pi}{n}}-e^{-\frac{ik\pi}{n}}+e^{-i\theta}\right) }$

$\small\text{ \displaystyle\longrightarrow \dfrac{e^{2in\theta}-1}{e^{2i\theta}-1}\cdot e^{i(1-n)\theta}=\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{\frac{ik\pi}{n}}-e^{-\frac{ik\pi}{n}}+e^{-i\theta}\right) }$

Then (1) becomes,

$\small\text{ \displaystyle\longrightarrow\dfrac{\sin(n\theta)}{\sin\theta}=\prod_{k=1}^{n-1}\left(e^{i\theta}-e^{\frac{ik\pi}{n}}-e^{-\frac{ik\pi}{n}}+e^{-i\theta}\right) }$

$\small\text{ \displaystyle\longrightarrow\dfrac{\sin(n\theta)}{\sin\theta}=2^{n-1}\prod_{k=1}^{n-1}\dfrac{\left(e^{i\theta}+e^{-i\theta}-e^{\frac{ik\pi}{n}}-e^{-\frac{ik\pi}{n}}\right)}{2} }$

$\small\text{ \displaystyle\longrightarrow\dfrac{\sin(n\theta)}{\sin\theta}=2^{n-1}\prod_{k=1}^{n-1}\left(\dfrac{e^{i\theta}+e^{-i\theta}}{2}-\dfrac{e^{\frac{ik\pi}{n}}+e^{-\frac{ik\pi}{n}}}{2}\right) }$

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