Probabilistic Proof of $\prod\limits_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\frac{\sin t}t$ 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.

Hint:
Let $b\in\mathbb{N}$, $b\ge2$, $Y=\{0,1,...,b-1\}$ with fair probabilities assigned, and let $F:Y^\infty\to[0,1]$ be the map defined by
$$F(x_1,x_2,x_3,...)=\frac{x_1}b+\frac{x_2}{b^2}+\frac{x_3}{b^3}+..., \text{for all}~(x_1,x_2,x_3,...)\in Y^\infty;$$
that is, $F=\sum_{n=1}^\infty\frac1{b^n}f_n$ where $f_n=\sum_{k=1}^{b-1}k\chi_{A_{nk}}$ and $A_{nk}=Y\times...\times Y\times\{k\}\times Y^\infty$
with ${k}$ in the $n$th spot. If $\mu:\mathcal{S}(\mathscr{C})\to[0,1]$ is the infinite product measure, then
$A\in\mathscr{B}$  if and only if $F^{-1}(A)\in\mathcal{S}(\mathscr{C})$, in which case $\mu(F^{-1}(A))=\mathfrak{m}(A)$.
Use this result and show that $f:[0,1]\to\overline{\mathbb{R}}$ is Borel
measurable if and only if $f\circ F:Y^\infty\to\overline{\mathbb{R}}$ is Borel measurable, and
$$\int_{[0,1]}f(x)dx=\int_{Y^\infty}f\circ F~d\mu\tag{*}$$
provided $f$ is nonnegative or integrable; in particular, it holds when $f:[0,1]\to\mathbb{C}$ is integrable.
Fix $z\in\mathbb{C}$. Then applying (*) to the function $f(x)=e^{2zx}$, prove that
$$e^z\frac{\sinh z}z=\lim_{N\to\infty}\int_{Y^\infty}\prod_{n=1}^N e^{2zf_n/b^n}d\mu$$
Write $\prod_{n=1}^N e^{2zf_n/b^n}=\prod_{n=1}^N\prod_{k=1}^{b-1}e^{2zk\chi_{A_{nk}}/b^n}$ as a simple function and use the
formula to compute $\int_{Y^\infty}\prod_{n=1}^N e^{2zf_n/b^n}d\mu$. Deduce the very interesting formula
$$e^z\frac{\sinh z}z=\prod_{n=1}^\infty\prod_{k=1}^{b-1}\left(1+\frac{2e^{kz/b^n}}b\sinh\left(\frac{kz}{b^n}\right)\right)$$
With $b=2$ and $z=it$, prove that
$$\frac{\sin t}t=\prod_{n=1}^\infty\cos\left(\frac t{2^n}\right)$$
