Recurrence Relation and finding cosine of a function of them.

Question

What if we are given $$a_{r+1}=\sqrt{\frac12(a_r+1)},r\in\{0\}\cup\mathbb N$$ How to find: $$\chi=\cos\left(\frac{\sqrt{1-a_0^2}}{\displaystyle\prod_{k=1}^{\infty}a_k}\right)$$ My try, let $a_0=1$ then $a_1=1,a_2=1,..$ then $\chi=\cos(0)=1$ but the answer is not 1.

Answer 1

Hint: recall the half-angle formula for cosine:

$$\cos{\frac{\theta}{2}} = \sqrt{\frac{1+\cos{\theta}}{2}} $$

Thus, if we let let $a_r = \cos{\theta_r}$, then $\theta_{r+1} = \theta_r/2$.