Are we in sinc?

Recall the double-angle identity for sines $\sin \theta = 2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$ .

Using the identity repeatedly on $\sin \frac{\theta}{2^{k}}$ gives the product

$\sin \theta = 2^{n} \sin \frac{\theta}{2^{n}} \prod_{k=1}^{n} \cos \frac{\theta}{2^{k}}$

for all positive integral $n$ . Taking the limit to infinity we have

$\lim_{n \rightarrow \infty} 2^{n} \sin \frac{\theta}{2^{n}} \prod_{k=1}^{n} \cos \frac{\theta}{2^{k}} = \lim_{n \rightarrow \infty} 2^{n} \frac{\theta}{2^{n}} \prod_{k=1}^{n} \cos \frac{\theta}{2^{k}} = \theta \prod_{k=1}^{\infty} \cos \frac{\theta}{2^{k}}$

since $\sin \theta \approx \theta$ when $\theta \rightarrow 0$ . This means that

$\prod_{k=1}^{\infty} \cos \frac{\theta}{2^{k}} = \frac{\sin \theta}{\theta}$

Substituting $\theta = \frac{\pi}{6}$ we finally get

$\prod_{k=1}^{\infty} \cos \frac{\theta}{2^{k}} = \frac{\sin \pi / 6}{\pi / 6} = \boxed{\frac{3}{\pi}}$

$\huge{\displaystyle \prod _{ k=0 }^{ n }{ \cos { \left( \frac { \theta }{ { 2 }^{ k } } \right) } } =\frac { \sin { \left( \theta \right) \cos { \left( \theta \right) } } }{ { 2 }^{ n }\sin { \left( \frac { \theta }{ { 2 }^{ n } } \right) } } }$

here $\theta=\frac{\pi}{12}$

$\large{\displaystyle \lim_{n\to \infty} \frac{\sin (\frac{\pi}{12}) \cos (\frac{\pi}{12})}{2^{n}\sin(\frac{\frac{\pi}{12}}{2^{n}})}}$

Now its easy to apply the limit, multiply and divide by $\frac{\frac{\pi}{12}}{2^{n}}$

we get finally

$\large{\frac{12}{\pi} \sin (\frac{\pi}{12})\cos (\frac{\pi}{12})=0.9549}$

I appreciate Jake Lai for his wonderful method.

Define $A_n = \cos\dfrac{\pi}{12} \cos\dfrac{\pi}{24}\cos\dfrac{\pi}{48}\cdots \cos\dfrac{\pi}{3\cdot 2^n} \cdot \sin\dfrac{\pi}{3\cdot 2^n}$

So,

$\begin{aligned} A_n &= \cos\dfrac{\pi}{12} \cos\dfrac{\pi}{24}\cos\dfrac{\pi}{48}\cdots \cos\dfrac{\pi}{3\cdot 2^n} \cdot \sin\dfrac{\pi}{3\cdot 2^n}\\ &=\dfrac{1}{2^{n-2}}\cos\dfrac{\pi}{12}\sin\dfrac{\pi}{12}\\ &=\dfrac{1}{2^{n-1}} \sin\dfrac{\pi}{6} \\ &= \dfrac{1}{2^n}\\ \text{Now, P is the} &\text{ required product, }\\ P &= \dfrac{A_n}{\sin \dfrac{\pi}{3\cdot 2^n}}\\ \lim_{x \to \infty} b_n &= \dfrac{1}{2^n} \cdot \dfrac{1}{\sin\dfrac{\pi}{3\cdot 2^n}}\\ \text{Now, when } n &\to \infty, ~ \frac{\pi}{3\cdot 2^n} \to 0 \\ \Rightarrow \lim_{x \to \infty} b_n &= \dfrac{1}{2^n}\dfrac{3\cdot 2^n} {\pi} \\ &= \boxed{\Large\dfrac{3}{\pi}}\\ &\approx 0.955 \end{aligned}$