How to prove $\lim_{n \to \infty} \cos \frac {\pi}{2^2}\cos \frac {\pi}{2^3}\cos \frac {\pi}{2^4}......\cos \frac {\pi}{2^n}=\frac {2}{\pi}$ I came across the following problem that says:   

prove that $$\lim_{n \to \infty} \cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......\cos \dfrac {\pi}{2^n}=\dfrac {2}{\pi}$$.  

My Attempt:  Let $$P=\lim_{n \to \infty} [\cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......\cos \dfrac {\pi}{2^n}] \implies \log P=\lim_{n \to \infty} \sum_{r=2}^{n}\log (\cos \dfrac {\pi}{2^r})$$. 
Now,I am stuck and not sure which way to go. Can  someone point me in the right direction?  Thanks in advance for your time.
 A: Let 
$$ g_n = \prod_{k=2}^n \cos \frac{\pi}{2^k}$$
Then
$$
    g_n \sin \frac{\pi}{2^n} = \frac{1}{2} g_{n-1} \sin \frac{\pi}{2^{n-1}}
$$
Meaning that
$$
   g_n \sin \frac{\pi}{2^n} = \frac{1}{2^{n-1}} g_1 = \frac{1}{2^{n-1}}
$$
Hence
$$
  g_n = \frac{2}{\pi} \frac{\frac{\pi}{2^{n}}}{\sin \frac{\pi}{2^{n}}} \longrightarrow_{n \to \infty} \frac{2}{\pi}
$$
A: Take the term inside the limit:
$$\cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......\cos \dfrac {\pi}{2^n}$$
Now multiply and divide by $2*\sin \dfrac {\pi}{2^n}$. You get:
$$\dfrac{\cos \dfrac {\pi}{2^2}\cos \dfrac {\pi}{2^3}\cos \dfrac {\pi}{2^4}......2*\sin \dfrac {\pi}{2^n}\cos \dfrac {\pi}{2^n}}{2*\sin \dfrac {\pi}{2^n}}$$
$2*\sin \dfrac {\pi}{2^n}\cos \dfrac {\pi}{2^n}$ = $\sin \dfrac {\pi}{2^{n-1}}$
Now multiply and divide by 2 and continue this process till you get :
$\dfrac{\sin \dfrac {\pi}{2}}{2^{n}*\sin \dfrac {\pi}{2^n}}$
Applying limit to this using the formula $$\lim_{x \to \infty}\dfrac{sin(x)}{x} = 1$$ will give you the answer as : $\dfrac{2}{\pi}$
A: Hint: Consider $\sin 2x=2\sin x\cos x$.
