I am interested to see if the product for the sinc function that is derived through the double angle formula: $$\frac{\sin(x)}{x}=\prod_{k=1}^\infty \cos\left(\frac{x}{2^k}\right)$$

can be manipulated into Euler's Product for the sinc: $$\frac{\sin(x)}{x}=\prod_{n=1}^\infty \left(1-\frac{x^2}{\pi^2 n^2}\right)$$

I attempted to introduce the taylor series for cosine but I am left with a complicated product of an inifnite sum and I do not know where to continue further.

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    $\begingroup$ Honestly speaking, I doubt that, unless we're speaking of a very wide sense of "manipulate". I mean, if you're just using the fact that $n=(2k-1)2^l$ and the Euler product of $\cos x$, it would be elementary, but I'd rather call that "cheating". $\endgroup$ – Professor Vector Jan 2 '18 at 15:44
  • $\begingroup$ I just want to see if I can derive it without resorting to Weierstrass factorization. $\endgroup$ – aleden Jan 2 '18 at 15:46
  • $\begingroup$ Hmm, you could try to show that both products have the same set of zeros and that they behave the same at $x=\infty$. then employing one of the mighty theorems of complex analysis (i forgot the name) would settle the case $\endgroup$ – tired Jan 2 '18 at 15:49
  • $\begingroup$ Well, if you want to do that without complex analysis, that's possible, but rather tedious and technical. You don't use just that $\sin2x=\sin x\cdot 2\cos x$ (as you did), but $\sin(n+1)x=\sin x\cdot U_n(\cos x)$, and the factorization of the polynomial $U_n$ due to its known zeros, and then you let $n\to\infty$... but it's not really that pretty. $\endgroup$ – Professor Vector Jan 2 '18 at 15:56

Hint: $$ \begin{align} \cos(x) &=\frac{\frac{\sin(2x)}{2x}}{\frac{\sin(x)}x}\\ &=\prod_{n=0}^\infty\left(1-\frac{4x^2}{(2n+1)^2\pi^2}\right) \end{align} $$

  • $\begingroup$ I just realized that this helps to prove the opposite direction. $\endgroup$ – robjohn Jan 2 '18 at 16:19

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