No

You can show by induction that $\displaystyle (x_n)_n$ is increasing, $\displaystyle (y_n)_n$ is decreasing and $\displaystyle \forall n~x_n \leq y_n$

Starting from this you can show that $\displaystyle (x_n)_n$ has a limit since it is increasing and majored by $\displaystyle y_0$. The same for $\displaystyle (y_n)_n$.

Then you can show that $\displaystyle (x_n)_n$ and $\displaystyle (y_n)_n$ are adjacent since $\displaystyle x_{n+1}- y_{n+1} = \frac{x_n-y_n}{2\:\left(1+\sqrt{\frac{2y_n}{x_n+y_n}}\right) } \leq \frac{x_n-y_n}{4}$

Then you can show that $\displaystyle (x_n)_n$ and $\displaystyle (y_n)_n$ have the same limit l.

You can see that $\displaystyle 2^n \arccos \frac{x_n}{y_n} = \arccos \frac{x_0}{y_0}$ may be true but $\displaystyle 2^n$ has infinite limit whereas $\displaystyle \arccos \frac{x_n}{y_n}$ converges towards $\displaystyle \arccos 1 = 0$