# 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

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