Is there a proof for the following fomula of Euler which does not use complex analysis or fourier analysis? $${\sin (\pi x)} = \pi x\prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2}\right).$$
Suppose that the student studied "Introduction to Real Analysis" by Bartle & Sherbert and they just learned what the convergence of an infinite product means. Then is there a proof that is suitable for them?
Let $I=[a,b]\subseteq\mathbb{R}$ and $\{f_n(x)\}_{n\in\mathbb{N}}$ the sequence of real polynomials defined by: $$ f_n(x) = \prod_{j=0}^{n}\left(1-\frac{4\,x^2}{(2j+1)^2\,\pi^2}\right). $$ We have that the sequence $\{f_n(x)\}_{n\in\mathbb{N}}$ is uniformly convergent to $\cos x$ over $I$.
From the factorization of Chebyshev polynomials of the first and second kind we have: $$ \frac{\sin x}{(2n+1)\,\sin\frac{x}{2n+1}}=\prod_{k=1}^{n}\left(1-\frac{\sin^2\frac{x}{2n+1}}{\sin^2\frac{k\pi}{2n+1}}\right),\qquad \cos x = \prod_{j=0}^{n-1}\left(1-\frac{\sin^2\frac{x}{2n}}{\sin^2\frac{(2j+1)\pi}{4n}}\right),$$ for any real number $x$ and for any $n\in\mathbb{N}^+$.
Without loss of generality we may assume $0 < x < m < n $ with $m$ and $n$ positive integers.
Since for any $\theta$ in the interval $\left(0,\frac{\pi}{2}\right)$ we have $\frac{2\theta}{\pi}<\sin\theta<\theta$, it follows that: $$\begin{eqnarray*}1&>&\prod_{k=m+1}^{n}\left(1-\frac{\sin^2\frac{x}{2n}}{\sin^2\frac{(2k+1)\pi}{4n}}\right)>\prod_{k=m+1}^{n}\left(1-\frac{x^2}{(2k+1)^2}\right)\\&>&1-x^2\!\!\sum_{k=m+1}^{n}\frac{1}{(2k+1)^2}>1-\frac{x^2}{4m},\end{eqnarray*}$$ so, if we take $$ H_m(x) = \prod_{j=0}^{m}\left(1-\frac{\sin^2\frac{x}{2n}}{\sin^2\frac{(2j+1)\pi}{4n}}\right), $$ we have that $\cos x$ belongs to the interval: $$ \left(\left(1-\frac{x^2}{4m}\right)\,H_m(x),\;\; H_m(x)\right). $$ By letting $n\to +\infty$ we have that $\cos x$ belongs to the interval: $$ \left[ \left(1-\frac{x^2}{4m}\right)\prod_{j=0}^{m}\left(1-\frac{4\,x^2}{(2j+1)^2\pi^2}\right),\;\; \prod_{j=0}^{m}\left(1-\frac{4\,x^2}{(2j+1)^2\pi^2}\right)\right], $$ then, letting $m\to +\infty$, the pointwise convergence of the Weierstrass product for the cosine function is proved. Additionally, from the last line it follows that: $$ \left| f_m(x)-\cos x \right| \leq \left| \cos x \right|\,\frac{4x^2/m}{1-4x^2/m} \leq \frac{4x^2}{m-4x^2}, $$ proving uniform convergence. Along the same lines, we can also prove the uniform convergence to $\sin x$ of the Weierstrass product for the sine function.
