Let $a_0 = 1$ and $a_n = a_{n-1}(\cos \frac{x}{2^n})$ .Find the limit $\{a_n\}$ when $x = \pi / 6$ . (i.e. find the $\lim_{n \to \infty} a_n$)
My try : I solved it but the solution was time-consuming .
$\frac{a_n}{a_{n-1}} = \cos \frac{x}{2^n} \to \frac{a_n}{a_0} = \frac{a_n}{a_{n-1}} \times \frac{a_{n-1}}{a_{n-2}} \times \dots \times\frac{a_1}{a_0} = \cos {\frac{x}{2^n}} \times \cos {\frac{x}{2^{n-1}}} \times \dots \times \cos {\frac{x}{2}} \to (2^n \sin{\frac{x}{2^n}})(a_n) = \sin x \to a_n = \frac{\sin x}{2^n\sin{\frac{x}{2^n}}} = \frac{\frac{x}{2^n}}{\sin{\frac{x}{2^n}}} \times \frac{\sin x }{x} \to \lim_{n \to \infty } a_n = 1 \times \frac{\sin{\frac{\pi}{6}}}{\frac{\pi}{6}} = \frac{3}{\pi} $
Note : I did some calculations by hand and didn't write them .
You already know that $$a_n = \cos \left({\frac{x}{2^n}}\right) \times \cos \left({\frac{x}{2^{n-1}}}\right) \times \dots \times \cos \left({\frac{x}{2}}\right) = \prod_{i=1}^n \cos \frac{\theta}{2^i} $$
Therefore $\displaystyle \lim_{n\to\infty} a_n = \prod_{i=1}^{\infty} \cos \dfrac{\theta}{2^i}$, if the product exists.
If you have only 1 minute to solve this, it is maybe expected that you know Viète's infinite product:
$$\prod_{n=1}^\infty \cos \frac{\theta}{2^n} = \frac{\sin \theta}{\theta}$$
Using this, we immediately get that the answer is $\dfrac{\sin (\pi/6)}{\pi/6} = \dfrac3{\pi}$.
So this is basically your solution, but with the addition that the second, perhaps more difficult step, is actually already a known identity. If you know the identity, you can do it in a minute.