We can see that $x_1 = 1 = 2 \sin \frac{\pi}{6}$, $y_1 = \sqrt{3} = 2 \cos \frac{\pi}{6}$.
So we will prove, by induction, that for all positive integers $n$, we have
$$x_n = 2 \sin \frac{\pi}{3 \cdot 2^n}, \quad y_n = 2 \cos \frac{\pi}{3 \cdot 2^n}. \qquad (1)$$
Indeed, for $n=1$ the statement is true. Assume that (1) is true for $n$. Applying the recurrent relation, we get
$$\begin{aligned} x_{n+1} &= \sqrt{2 - y_n} = \sqrt{2 - 2 \cos \frac{\pi}{3 \cdot 2^n}} \\ &= \sqrt{4 \sin^2 \frac{\pi}{3 \cdot 2^{n+1}}} = 2 \sin \frac{\pi}{3 \cdot 2^{n+1}} \end{aligned}$$
and
$$y_{n+1} = \frac{x_n}{x_{n+1}} = \frac{2 \sin \frac{\pi}{3 \cdot 2^n}}{2 \sin \frac{\pi}{3 \cdot 2^{n+1}}} = 2 \cos \frac{\pi}{3 \cdot 2^{n+1}}.$$
So (1) is also true for $n+1$. Hence, (1) is true for all positive integers $n$. From this we have
$$\lim x_n = \lim \left( 2 \sin \frac{\pi}{3 \cdot 2^n} \right) = (2 \sin 0) = 0$$
and
$$\lim y_n = \lim \left( 2 \cos \frac{\pi}{3 \cdot 2^n} \right) = (2 \cos 0) = 2.$$
Hence, $(x_n)$, $(y_n)$ are convergence and $\lim x_n = 0$, $\lim y_n = 2$. $\square$