sequences and series - Convergence of $a_{n+1}=sqrt{2-a_n}$

I'm attempting to prove the convergence (or divergence, though I strongly suspect it converges) of a sequence defined as $a_{n+1}=\sqrt{2-a_n}$ with $a_1=\sqrt{2}$.

I cannot use the monotonic sequence theorem as the sequence is not monotonically increasing. In fact, the first few values of the sequence are:

$a_1 =\sqrt{2}\approx 1.4142$

$a_2 =\sqrt{2-\sqrt{2}}\approx .7653$

$a_3 =\sqrt{2-\sqrt{2-\sqrt{2}}}\approx 1.1111$

Thus, it seems that $a_{n \to \infty} \to 1$

It seems that the sequence is behaving similarly to $\frac{\sin x}{x}$, leading me to think that the squeeze theorem may be useful. Still, I cannot seem to make any progress besides numerical computation of successive terms.

Answer

Hint: write $a_n=1+b_n$ or $a_n=1-b_n$, whichever makes $b_n$ positive. How does $b_n$ behave?

Elaboration: we have $a_n=1+b_n$ for odd $n$ and $a_n=1-b_n$ for even $n$ (why so?). So, for example, for even $n$ we can write $a_{n+1}=\sqrt{2-a_n}$ as $1+b_{n+1}=\sqrt{2-(1-b_n)}=\sqrt{1+b_n}$. Now you can compare $b_{n+1}$ and $b_n$. Proceed similarly for odd $n$.