How to find the value of $\dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{2 + \sqrt{2}}}{2}\cdot \dfrac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2}...$?
I can notice that, $a_{n-1} = 2(a_n)^2 -1$
This will further imply that, $a_n = \sqrt{\dfrac{a_{n-1} + 1}{2}}$
I'm thinking of relating this general term with the cosine identity. $\cos(\frac{x}{2})= \sqrt{\frac{1 +\cos x}{2}}$
Also I can notice that, $a_1 = \dfrac{2\cos(\frac{\pi}{4})}{2}$
Can we conclude anything from the above points?
You found that $$a_{n+1}=\sqrt{\frac{a_n+1}{2}}$$ and $a_1=\cos\frac{\pi}{4}$
Let $x=\frac{\pi}{4}$
Therefore using this the product becomes
$$P=\cos x \cos \frac{x}{2}\cos\frac{x}{4}\cos \frac{x}{8}...$$
We can calculate $P$ by taking limiting value of $P_n$where
$$P_n=\cos x \cos \frac{x}{2}\cos\frac{x}{4}\cos \frac{x}{8}...\cos\dfrac{x}{2^n}=\frac{\sin2x}{2^{n+1}\sin\frac{x}{2^n}}$$
$$P=\lim_{n \to \infty}P_n=\lim_{n\to\infty}\frac{\sin(\frac{\pi}{4}*2)}{2^{n+1}\sin\frac{\pi}{2^{n+2}}}=\frac{2}{\pi}$$