$b_{n}$ is increasing I think there is misunderstanding in my last post because its contain three questions so i will post question by question step by step 
An inequality for the product $\prod_{k=2}^{n}\cos\frac{\pi }{2^{k}}$

Let $n\geq 2\quad a_{n}=\prod\limits_{k=2}^{n}\cos\left(\dfrac{\pi }{2^{k}}\right)$ and $b_{n}=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$

Show that $b_{n}$ is increasing 
my attempts :


*

*$b_{n+1}-b_{n}=a_{n+1}\cos\left(\dfrac{\pi }{2^{n+1}}\right)-a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)=a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)\cos\left(\dfrac{\pi }{2^{n+1}}\right)-a_{n}\cos\left(\dfrac{\pi }{2^{n}}\right)$

*$\dfrac{b_{n+1}}{b_n}=\cos(\pi/2^{n+1})$


im stuck here in these two ways
any help would be appreciated

Notice that $\dfrac{b_{n+1}}{b_n}=\dfrac{\cos^2\left(\dfrac{\pi}{2^{n+1}}\right)}{\cos\left(\dfrac{\pi}{2^n}\right)}=\dfrac{\dfrac{1+\cos\left(\dfrac{\pi}{2^n}\right)}{2}}{\cos\left(\frac{\pi}{2^n}\right)}=\dfrac{1+\cos\left(\dfrac{\pi}{2^n}\right)}{2\cos\left(\dfrac{\pi}{2^n}\right)}>\dfrac{2\cos\left(\dfrac{\pi}{2^n}\right)}{2\cos\left(\dfrac{\pi}{2^n}\right)}=1$.
