$b_{n}$ 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

A variant: $$\begin{align*} b_n <b_{n+1} &\iff \cos\Bigl(\frac{\pi}{2^n}\Bigr) < \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)\\ &\iff 2\cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)-1 < \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr)\\ &\iff \cos^2\Bigl(\frac{\pi}{2^{n+1}}\Bigr) <1 .\end{align*}$$
