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.

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Consider the following variable
$$
Y=\sum_{j=1}^{\infty}X_j2^{-j}
$$
where $X_i$’s are i.i.d. unbiased Bernoulli$\{-1,1\}$. The characteristic function of the right hand side turns out to be
$$
\prod_{j=1}^{\infty}\cos(t2^{-j})
$$
On the other hand you can convince yourself that $Y$ is Uniform$(-1,1)$ (I am not doing this part of the proof, but you can think of Uniform measure as a ”uniform” binary expansion.)
Then the characteristic function of $Y$ is
$$
\int_{-1}^{1}\frac{e^{itx}}{2}dx=\frac{e^{it}-e^{-it}}{2it}=\frac{\sin (t)}{t}
$$
and we are done.

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
$$\frac{\sin t}t=\prod_{n=1}^\infty\cos\left(\frac t{2^n}\right)$$

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
$$ \prod_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\lim_{n\to\infty}\prod_{i=1}^n\cos\left(\frac t{2^i}\right)=\lim_{n\to\infty}\frac{\sin t}{2^{n}\sin\left(\frac t{2^n}\right)}=\frac{\sin t}{t}.$$