More Elementary Proof for Euler’s Sine Expansion

Our goal in this post is to provide an elementary proof for the real version of Euler’s famous Sine expansion.

I first prove that for any n that is a power of 2 and for any x\in\mathbb R,

\displaystyle{\sin(x)=n p_n(x)\sin\left(\frac x n\right)\cos\left(\frac x n\right),~p_n(x)=\prod_{k=1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)}~~~~~(1)

by using the following more familiar identity

\displaystyle\sin(x)=2\sin\frac x 2\sin\frac{\pi+x}2

we can easily arrive to the following identity that is valid for n equal to any power of 2,

                    \displaystyle\sin(x)=2^{n-1}\prod_{k=0}^{n-1}\sin\frac{k\pi+x}n

\displaystyle=2^{n-1}\sin\frac x n\sin\frac{\frac n 2\pi+x}n\prod_{k=1}^{\frac n 2-1}\sin\frac{k\pi+x}n\sin\frac{k\pi-x}n

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

By considering what happens as x\to0 in the recent formula, we can obtain that for n a power of 2

\displaystyle n=2^{n-1}\prod_{k=1}^{\frac n 2-1}\sin^2\frac{k\pi}n~~~~~~~~~~(3)

and then I replace (3) in (2) and prove (1).

Now I choose m<\frac n 2-1 and break up p_n(x) as follows

\displaystyle p_n(x)=\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)~~~~~~~~~~(4)

Since for 0<\theta<\frac{\pi}2, \sin(\theta)>\frac2{\pi}\theta, we have

\displaystyle\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{x^2}{4k^2}

and if I choose m and n large enough such that \frac{x^2}{4m^2}<1, then

\displaystyle0<1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<1,k=m+1,...,\frac n 2-1

which implies that

\displaystyle0<\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)<1

Therefore from (4) we have

\displaystyle p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)

Also we have

\displaystyle{\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\ge1-\sum_{k=m+1}^{\frac n 2-1}\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\ge1-\frac{x^2}4\sum_{k=m+1}^\infty\frac1{k^2}=1-s_m}

and once again by (4), we get

\displaystyle p_n(x)\ge(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)

To summarize, I have shown that

\displaystyle{(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\le p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)}~~~~~~~~~~(5)

Now taking n\to\infty in (5) I arrive to

\displaystyle(1-s_m)\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\le\frac{\sin(x)}x\le\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)

After rearrangement and then taking absolute values, we get

\displaystyle\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|

but since we have

\displaystyle{\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le\prod_{k=1}^m\left(1+\frac{x^2}{k^2\pi^2}\right)\le\prod_{k=1}^me^{\frac{x^2}{k^2\pi^2}}\le e^{\sum_{k=1}^\infty\frac{x^2}{k^2\pi^2}}=e^{\ell}}

I can write

\displaystyle\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|e^{\ell}

Finally since \lim_{m\to\infty}s_m=0, this inequality implies our goal, indeed

\boxed{\displaystyle\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)=\frac{\sin(x)}x}

More Elementary Proof for Euler’s Sine Expansion

One Finite Sum

In this post I want to find the value of the following finite sum,

\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}

I use the following amazing identity that you can see it’s proof in my previous posts

\displaystyle\frac{1}{\sin^2(x)}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}

By this identity, I can write

                    \displaystyle\sum_{k=0}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(\frac{x+k\pi}{m}+n\pi)^2}

                                                           \displaystyle=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{\frac{(x+k\pi+mn\pi)^2}{m^2}}

                                                          \displaystyle=m^2\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(x+k\pi+mn\pi)^2}

                                                          \displaystyle=m^2\sum_{n\in\mathbb{Z}}\sum_{k=0}^{m-1}\frac{1}{(x+(k+mn)\pi)^2}=\frac{m^2}{\sin^2(x)}

and this follows that

\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}

Hence,

\boxed{\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}=\lim_{x\to0}\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}=\frac{m^2-1}{3}}

One Finite Sum