calculus - $dfrac{2}{pi} = dfrac{sqrt 2}{2} cdot dfrac{sqrt {2+sqrt 2}}{2} cdotdfrac{sqrt {2+sqrt {2+sqrt 2}}}{2} cdots$

One can show inductively that
$$\cos \frac{\pi}{2^{n+1}}\ = \frac{\sqrt {2+\sqrt {2+\sqrt {2+\sqrt {\cdots+\sqrt {2 }}}}}}{2},$$
with $n$ square roots in the right side of the equation.

The second part of the question was to deduct the following from the first part:

$$\frac{2}{\pi} = \frac{\sqrt 2}{2} \cdot \frac{\sqrt {2+\sqrt 2}}{2} \cdot\frac{\sqrt {2+\sqrt {2+\sqrt 2}}}{2} \cdot \cdots$$

with the hint to use the following limit:

$$\lim_{n\to \infty}\cos\Big(\frac{t}{2}\Big)\cos\Big(\frac{t}{2^2}\Big)\cdots\cos\Big(\frac{t}{2^n}\Big) = \frac{\sin t}{t}.$$

A hint or some general intuition will be appreciated.

Answer

Using the trogonometric identity
$$\sin (2a)=2\sin a\,\cos a\qquad\text{or}\qquad \cos a=\frac{\sin 2a}{2\sin a},$$
provided that $\,\sin a\ne 0,\,$ we obtain that
$$\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{2\sin(x/2)}\frac{\sin (x/2)}{2\sin(x/4)}\cdots\frac{\sin (x/2^{n-1})}{2\sin(x/2^n)}=\frac{\sin x}{2^n\sin(x/2^n)}.$$
Hence

$$\lim_{n\to\infty}\cos(x/2)\cos(x/4)\cdots\cos(x/2^n)=\frac{\sin x}{x},$$
since
$$\lim_{t\to 0}\frac{\sin(tx)}{t}=x.$$
In particular,
$$\prod_{n=1}^\infty \cos\left(\frac{\pi}{2^{n+1}}\right)=\frac{2}{\pi}.$$