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I've been looking at proofs of Euler's sine expansion, that is $$\frac{\sin x}{x}=\prod_{k=1}^\infty\left({1-\frac{x^2}{k^2\pi^2}}\right)$$

All the proofs seem to rely on complex analysis and Fourier series.

Is there any more elementary proof?

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I first prove that for any $n$ that is a power of $2$ and for any $x\in\mathbb R$, $$\sin(x)=n p_n(x)\sin\left(\frac x n\right)\cos\left(\frac x n\right),~p_n(x)=\prod_{k=1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{1}$$ using the following more familiar identity $$\sin(x)=2\sin\frac x 2\sin\frac{\pi+x}2$$ we can easily arrive to the following identity that is valid for $n$ equal to any power of $2$, $$\begin{align}\sin(x)&=2^{n-1}\prod_{k=0}^{n-1}\sin\frac{k\pi+x}n\\ &=2^{n-1}\sin\frac x n\sin\frac{\frac n 2\pi+x}n\prod_{k=1}^{\frac n 2-1}\sin\frac{k\pi+x}n\sin\frac{k\pi-x}n\\ &=2^{n-1}\sin\frac x n\cos\frac x n\prod_{k=1}^{\frac n 2-1}\left(\sin^2\frac{k\pi}n-\sin^2\frac x n\right)\tag{2}\end{align}$$ By considering what happens as $x\to0$ in the recent formula, we can obtain that for $n$ a power of $2$ $$n=2^{n-1}\prod_{k=1}^{\frac n 2-1}\sin^2\frac{k\pi}n\tag{3}$$ 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 $$p_n(x)=\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{4}$$ Since for $0<\theta<\frac{\pi}2$, $\sin(\theta)>\frac2{\pi}\theta$, we have $$\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{x^2}{4k^2}$$ and if I choose $m$ and $n$ large enough such that $\frac{x^2}{4m^2}<1$, then $$0<1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<1,k=m+1,...,\frac n 2-1$$ which implies that $$0<\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)<1$$ Therefore from $(4)$ we have $$p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$$ Also we have $$\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\ge1-\sum_{k=m+1}^{\frac n 2-1}\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\ge1-\frac{x^2}4\sum_{k=m+1}^\infty\frac1{k^2}=1-s_m$$ and once again by $(4)$, we get $$p_n(x)\ge(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$$ To summarize, I have shown that $$(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\le p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\tag{5}$$ Now taking $n\to\infty$ in $(5)$ I arrive to $$(1-s_m)\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\le\frac{\sin(x)}x\le\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)$$ After rearrangement and then taking absolute values, we get $$\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|$$ but since we have $$\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le\prod_{k=1}^m\left(1+\frac{x^2}{k^2\pi^2}\right)\le\prod_{k=1}^me^{\frac{x^2}{k^2\pi^2}}\le e^{\sum_{k=1}^\infty\frac{x^2}{k^2\pi^2}}=e^{\ell}$$ I can write $$\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|e^{\ell}$$ Finally since $\lim_{m\to\infty}s_m=0$, this inequality implies our goal, indeed $$\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)=\frac{\sin(x)}x$$

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    $\begingroup$ Thanks, I figured it out. It's just flipped. $\endgroup$ – Superbus May 8 '14 at 14:40
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    $\begingroup$ Shouldn't (6) be $\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{n^2}{4k^2}$? That's what I get using the inequality. $\endgroup$ – Superbus May 8 '14 at 15:08
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    $\begingroup$ Note that $\sin^2\frac x n<\frac{x^2}{n^2}$. $\endgroup$ – user91500 May 8 '14 at 15:11
  • $\begingroup$ Nice proof, but I'm not fully convinced by (8) and (9). What if $$\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)$$ is negative? And what if $1 - s_m$ is negative? $\endgroup$ – user85798 May 12 '14 at 11:30
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    $\begingroup$ @Superbus In the following paper, there is another similar proof jstor.org/stable/2690668?origin=JSTOR-pdf $\endgroup$ – user91500 May 14 '14 at 16:10
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Here's a proof using elementary (albeit not very well known) trigonometric integrals.

We may assume $0 < x \leq 1/2$ because the ratio of $\sin(\pi x) / (\pi x)$ to $\prod_{k=1}^\infty \big(1 - (x/k)^2\bigl)$ is readily seen to be an even function of period $1$.

For $n \geq 0$ and $c \in {\bf R}$ define $$ I_n(c) := \int_0^{\pi/2} \cos^n t \, \cos ct \, dt. $$ Then $I_0(0) = \pi/2$ and $I_0(2x) = \sin(\pi x) / (2x)$. Hence $$ \frac{I_0(2x)}{I_0(0)} = \frac{\sin \pi x}{\pi x}. $$ I claim that also $$ \frac{I_0(2x)}{I_0(0)} = \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2} \right), $$ which will prove the desired product formula.

The key is a recursion for the ratios $I_n(2x) / I_n(0)$ obtained from the following integration by parts:

Lemma. For $n \geq 2$, $$ (n^2 - c^2) I_n(c) = (n^2-n) I_{n-2}(c). $$

Proof: Integrate by parts twice to find $$ c^2 I_n(c) = c \int_0^{\pi/2} \cos^n t \, d(\sin ct) = n c \int_0^{\pi/2} \cos^{n-1} t \, \sin t \, \sin ct \, dt $$ $$ = -n \int_0^{\pi/2} \cos^{n-1} t \, \sin t \, d(\cos ct) = n \int_0^{\pi/2} (\cos^n t - (n-1) \cos^{n-2} t \sin^2 t) \, \cos ct \, dt. $$ But $\sin^2 t = 1 - \cos^2 t$, so $$ c^2 I_n(c) = n \int_0^{\pi/2} (n \cos^n t - (n-1) \cos^{n-2} t) \, \cos ct \, dt = n^2 I_n(c) - (n^2-n) I_{n-2}(c), $$ and the recursion follows. $\Box$

Therefore $$ \frac{I_{n-2}(c)}{I_{n-2}(0)} = \frac{n^2-c^2}{n^2} \frac{I_n(c)}{I_n(0)}, $$ whence by induction on $m=1,2,3,\ldots$ $$ \frac{I_0(c)}{I_0(0)} = \prod_{k=1}^m \frac{(2k)^2-c^2}{(2k)^2} \cdot \frac{I_{2m}(c)}{I_{2m}(0)}. $$ Taking $c=2x$ we find $$ \frac{\sin \pi x}{\pi x} = \prod_{k=1}^m \frac{k^2-x^2}{k^2} \cdot \frac{I_{2m}(2x)}{I_{2m}(0)} = \prod_{k=1}^m \left( 1 - \frac{x^2}{k^2} \right) \cdot \frac{I_{2m}(2x)}{I_{2m}(0)}, $$ and it remains to prove that $I_n(2x) / I_n(0) \rightarrow 1$ as $n \rightarrow \infty$. But this follows from the inequalities $$ I_n(0) > I_n(2x) > I_{n+2}(0) $$ (the second inequality is where we use $x \leq 1/2$: we need $\cos 2xt \leq \cos^2 x$), since by our Lemma $I_{n+2}(0) / I_n(0) = (n+1)/(n+2)$, which approaches $1$ as $n \rightarrow \infty$. Taking $m \rightarrow \infty$ we deduce $$ \frac{\sin \pi x}{\pi x} = \prod_{k=1}^\infty \left( 1 - \frac{x^2}{k^2} \right), $$ QED.

I noticed this argument some years back (see for instance #8 of this problem set from 2000). Naturally the formula for integrals $I_n(c)$ is known (even for non-integer $n$, see integral 3.631#9 in Gradshteyn and Ryzhik), but I don't know if and where it was previously noticed that one can use these integrals to prove the product formula for the sine.

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I think this is the most motivated elementary proof.

For $z \in \mathbb C$ and $n \in \mathbb N_{> 0}$, let:

$$f_n \left({z}\right) = \frac 1 2 \left[{\left({1 + \frac z n}\right)^n - \left({1 - \frac z n}\right)^n }\right]$$

Then $f_n\left({z}\right) = 0$ if and only if:

\begin{align} &&\left({1 + \frac z n}\right)^n & = \left({1 - \frac z n}\right)^n \\ & \iff & 1 + \frac z n & = \left({1 - \frac z n}\right) e^{2 \pi i \frac k n} \\ & \iff & z & = n \frac {e^{2 \pi i \frac k n} - 1} {e^{2 \pi i \frac k n} + 1} \\ &&& = n i \tan \left({\frac {k \pi} n }\right) \end{align}

Let $n = 2 m + 1$.

Then the roots of $f_{2 m + 1} \left({z}\right)$ are $\left({2 m + 1}\right) i \tan \left({\dfrac {k \pi} {2 m + 1}}\right)$ for $- m \le k \le m$.

Observe that $f_{2m + 1} \left({z}\right)$ is a polynomial of degree $2 m + 1$.

Then for some constant $C$, we have:

\begin{align} f_{2 m + 1} \left({z}\right) & = C z \prod_{\substack {k \mathop = - m \\ k \mathop \ne 0} }^m \left({1 - \frac z {\left({2 m + 1}\right) i \tan \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)\\ & = C z \prod_{k \mathop = 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \end{align}

It can be seen from the binomial theorem that the coefficient of $z$ in $f_n \left({z}\right)$ is $1$.

Hence $C = 1$, and we obtain:

$$f_{2 m + 1} \left({z}\right) = z \prod_{k \mathop = 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)$$

First we consider $z = x$ where $x$ is a non-negative real number.

Let $l < m$.

Then:

$$x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \le f_{2 m + 1} \left({x}\right)$$

Taking the limit as $m \to \infty$ we have:

\begin{align} & &\lim_{m \to \infty} x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} \left({\frac {k \pi / \left({2 m + 1}\right)} {\tan \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)^2 }\right) & \le \frac 1 2 \left({e^x - e^{- x} }\right)\\ & \implies &x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) & \le \sinh x \end{align}

By the inequality $\tan \theta \ge \theta$ for $0 \le \theta < \dfrac {\pi} 2$ we have:

$$f_{2 l + 1} \left({x}\right) \le x \prod_{k \mathop = 1}^l \left({1 + \frac {x^2} {k^2 \pi^2} }\right) \le \sinh x$$

Taking the limit as $l \to \infty$ we have by Squeeze Theorem:

$$\quad x \prod_{k \mathop = 1}^\infty \left({1 + \frac {x^2} {k^2 \pi^2} }\right) = \sinh x \tag{1}$$

Now let $1 < l < m$.

We have:

\begin{align} &\left \vert{f_{2 m + 1} \left({z}\right) - z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right \vert \\ & =\left \vert{z}\right \vert \left \vert{\prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right \vert \cdot \left \vert{\prod_{k \mathop = l + 1}^m \left({1 + \frac {z^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) - 1}\right \vert\\ & \le \left \vert{z}\right \vert \left[{\prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right)}\right] \cdot \left[{\prod_{k \mathop = l + 1}^m \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) - 1}\right] \\ & = f_{2 m + 1} \left({\left \vert{z}\right \vert}\right) - \left \vert{z}\right \vert \prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {\left({2 m + 1}\right)^2 \tan^2 \left({k \pi / \left({2 m + 1}\right)}\right)} }\right) \end{align}

Taking the limit as $m \to \infty$ we have:

$$\left \vert{\sinh z - z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {k^2 \pi^2} }\right)}\right \vert \le \sinh {\left \vert{z}\right \vert} - \left \vert{z}\right \vert \prod_{k \mathop = 1}^l \left({1 + \frac {\left \vert{z}\right \vert^2} {k^2 \pi^2} }\right)$$

Now take the limit as $l \to \infty$.

By $(1)$ and Squeeze Theorem, we have:

$$\sinh z = z \prod_{k \mathop = 1}^l \left({1 + \frac {z^2} {k^2 \pi^2} }\right)$$

Finally, substituting $z \mapsto i z$, we obtain:

$$ \sin z = z \prod_{k \mathop = 1}^l \left({1 - \frac {z^2} {k^2 \pi^2} }\right)$$

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In case you want an extremely simple proof:

Let's rewrite the product formula as follows:

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

Note that the right-hand side of the equation is a polynomial of a degree of $\infty$, factored into a product of other smaller polynomials of a degree of 2.

Now, the equation holds if the roots of the polynomial and the roots of the sine function coincide. Roots of the sine function:

$$x=0,\pm\pi,\pm2\pi,... $$

In similar manner, you can write down a product formula for the $\cos x$ if you know that the roots of the cosine are

$$\pm\frac{2k-1}{2}\pi;\;k=0,1,2,... $$

etc.

Remark:

"First of all, the result, and then rigor if needed."

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  • $\begingroup$ This is all the proof I need. $\endgroup$ – Kainui Sep 22 '14 at 16:01
  • $\begingroup$ This isn't really a proof, but it gives a good intuition why it works. $\endgroup$ – Rufflewind Sep 27 '14 at 8:59

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