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Would appreciate some ideas for the following:

"Prove that $\frac{\sin{x}}{x}=\prod_{n=1}^{\infty}\cos{\frac{x}{2^n}}$ using power series."

I'm aware this identity can be shown using trig identities and a telescoping product. Also, you can get another proof using the infinite product expressions $\sin{x} = x\prod_{k=1}^{\infty} (1-\frac{x^2}{k^2 \pi^2})$ and $\cos{x}=\prod_{k=1, \ k \ \text{odd}}^\infty (1-\frac{4x^2}{k^2 \pi^2})$.

However, since the question explicitly mentions power series, I was wondering if there is a proof that directly uses power series? I've tried calculating some derivatives and coefficients but they seem to get pretty nasty.

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  • $\begingroup$ use Taylor series for sin at zero? $\endgroup$ – Dole Dec 23 '18 at 6:26
  • $\begingroup$ yeah, but the RHS is giving me trouble $\endgroup$ – Merk Zockerborg Dec 23 '18 at 6:28
  • $\begingroup$ Is there a nice power series for $\log(\cos x)$? $\endgroup$ – mathworker21 Dec 23 '18 at 6:44
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Using the fact that $\sin(2x)=2\sin(x)\cos(x)$ we have $\cos(x)=\frac{\sin(2x)}{2\sin(x)}$.

Define $p_n(x):=\prod\limits_{j=1}^n \cos(\frac{x}{2^j})$ and note that $\cos(\frac{x}{2^j})=\dfrac{\sin(\frac{x}{2^{j-1}})}{2\sin(\frac{x}{2^j})}$ for each $j$.

Now, $p_n(x)=\frac{1}{2^n}\frac{\sin(x)}{\sin(\frac{x}{2^n})}=\frac{\sin(x)}{x}\dfrac{\frac{x}{2^n}}{\sin(\frac{x}{2^n})}$ and $p_n(x)\to_n \frac{\sin(x)}{x}$

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HINT

$\cos z= e^{iz}(1+e^{-2iz})/2$ so $ln \cos z=\cdots$

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