If $a$, $b$ are positive quantities such that ($a < b$) and if $$a_1=\frac{a+b}{2},b_1=\sqrt{a_1b},\ a_2=\frac{a_1+b_1}{2},b_2=\sqrt{a_2b_1},\ldots,\\ a_n=\frac{a_{n-1}+b_{n-1}}{2},b_n=\sqrt{a_{n}b_{n-1}}$$
Prove that $$\lim_{n \to \infty}b_n=\dfrac{\sqrt{b^2-a^2}}{\cos^{-1}(b/a)}$$
I tried to show that as $n\to \infty,a_n=b_n\implies a_{n-1}=b_{n-1}$
But I do not find this result quite promising. Please help. Some hint/solution is appreciated.
Hint
You have $$\begin{cases} a_n=\frac{a_{n-1}+b_{n-1}}{2}\\ b_n=\sqrt{a_nb_{n-1}} \end{cases}$$
The second equation gives $a_n=\frac{b_n^2}{b_{n-1}}$. Replace this quantity in the first one, set $z_n=\frac{b_n}{b_{n-1}}$ and see what happen.