TITLE: Probabilistic Proof of $\prod\limits_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\frac{\sin t}t$
QUESTION [12 upvotes]: Using probability methods (characteristic function?) prove 
$$\prod_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\frac{\sin t}t$$
I know what is characteristic function but I have no idea how use it in this task. I will grateful for help.

REPLY [3 votes]: I don't know if it answers your question since I don't use characteristic function. By using
$$ \sin(2\theta)=2\sin\theta\cos\theta $$
one has
\begin{eqnarray}
\prod_{i=1}^n\cos\left(\frac t{2^i}\right)&=&\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)\cos\left(\frac t{2^n}\right)\\
&=&\frac{\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)\cos\left(\frac t{2^n}\right)\sin\left(\frac t{2^n}\right)}{\sin\left(\frac t{2^n}\right)}\\
&=&\frac{\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)2\cos\left(\frac t{2^n}\right)\sin\left(\frac t{2^n}\right)}{2\sin\left(\frac t{2^n}\right)}\\
&=&\frac{\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)\sin\left(\frac t{2^{n-1}}\right)}{2\sin\left(\frac t{2^n}\right)}
\end{eqnarray}
and continue doing this, one has
\begin{eqnarray}
\prod_{i=1}^n\cos\left(\frac t{2^i}\right)
&=&\cdots\\
&=&\frac{\cos\left(\frac t{2}\right)\sin\left(\frac t{2}\right)}{2^{n-1}\sin\left(\frac t{2^n}\right)}
&=&\frac{\sin t}{2^{n}\sin\left(\frac t{2^n}\right)}.
\end{eqnarray}
Thus
$$ \prod_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\lim_{n\to\infty}\prod_{i=1}^n\cos\left(\frac t{2^i}\right)=\lim_{n\to\infty}\frac{\sin t}{2^{n}\sin\left(\frac t{2^n}\right)}=\frac{\sin t}{t}.$$