$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$ Given
$$a_n=\sqrt{2-a_{n-1}},a_1=\sqrt{2}$$
I calculated $a_1$ to $a_5$
$$\sqrt{2},
\sqrt{2-\sqrt{2}},
\sqrt{2-\sqrt{2-\sqrt{2}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}, \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$
which made me think of $\sin/\cos$. So I divided each by $2$, calculated $\arcsin(x)$ and got $$\left\{\frac{\pi }{4},\frac{\pi }{8},\frac{3 \pi }{16},\frac{5 \pi }{32},\frac{11 \pi }{64}\right\}$$
I found that this could be a formula
$$\frac{\pi}{3} \cdot  (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}$$
So $$\sin \left(\frac{\pi}{3} \cdot  (-1)^n \cdot\frac{(-2)^n-1}{2^{n+1}}\right)$$  is $a_n/2$, but this way seems not natural. Is there a more natural way?

If the question is about finding the limit, let's consider $a_{n+1}=f(a_n)$, where $f(x)=\sqrt{2-x}$. Then we have

If $0\leq x \leq \sqrt{2}$ then $0\leq f(x)\leq\sqrt{2}$

Indeed $0\leq x\leq\sqrt{2} 
\Rightarrow 0\geq-x \geq -\sqrt{2} 
\Rightarrow 2\geq 2-x \geq 2-\sqrt{2}>0
\Rightarrow \sqrt{2}\geq \sqrt{2-x}=f(x)\geq 0$.

Now, let's use Banach fixed-point theorem, and MVT, given $f'(x)=-\frac{1}{2\sqrt{2-x}}$, for $\forall x,y \in[0,\sqrt{2}], x<y$, there $\exists\varepsilon\in (x,y)$ s.t.
$$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|=
\frac{1}{2\sqrt{2- \varepsilon}}\cdot|x-y|<
\frac{1}{2\sqrt{2-\sqrt{2}}}\cdot|x-y|$$
since $\varepsilon\in[0,\sqrt{2}]$ as well. It's not to difficult to check that $0<\frac{1}{2\sqrt{2-\sqrt{2}}}<1$.
So, the limit exists and you can legitimately use $L=\sqrt{2-L}$ to find it, considering that $L\in[0,\sqrt{2}]$ of course, since all $(a_n)_{n>0} \subset[0,\sqrt{2}]$. 

Remark: More interesting results here. 

The trick that is typically applied for the $+$ with $\cos$ may not easily apply for $\arcsin$ and $\sin$ since:
$$\sin{\frac{\pi}{8}}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$
$$\sin{\frac{\pi}{16}}=\frac{1}{2}\sqrt{2-\sqrt{2\color{red}{+}\sqrt{2}}}$$
I'd rather try induction, given that
$$\sin{\frac{\color{red}{1}\cdot\pi}{4}}=\frac{\sqrt{2}}{2}$$
$$\sin{\frac{\color{red}{1}\cdot\pi}{8}}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$
$$\sin{\frac{\color{red}{3}\cdot\pi}{16}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2}}}$$
$$\sin{\frac{\color{red}{5}\cdot\pi}{32}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}$$
$$\sin{\frac{\color{red}{11}\cdot\pi}{64}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}$$
$$\sin{\frac{\color{red}{21}\cdot\pi}{128}}=\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}}$$
where $\{1,1,3,5,11,21\}$ is the begining of the Jacobsthal sequence, according to OEIS.
And of course: 
$$\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)=\frac{1}{2}\left(1-\sin{x}\right)\tag{1}$$
Jacobsthal sequence is $J_{n+1}=2^n-J_n$ and, assuming induction hypothesis, we have
$$\sqrt{\frac{1}{2}\left(1-\sin\left(\color{red}{J_n}\frac{\pi}{2^{n+1}}\right)\right)}=\\
\sqrt{\frac{1}{2}\left(1-\frac{1}{2}\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}\right)}=\\
\sqrt{\frac{1}{4}\left(2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}\right)}=\\
\frac{1}{2}\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}\overset{(1)}{=}\\
\sin\left(\frac{\pi}{4}-J_n\frac{\pi}{2^{n+2}}\right)=
\sin\left(\left(2^n-J_n\right)\frac{\pi}{2^{n+2}}\right)=
\sin\left(\color{red}{J_{n+1}}\frac{\pi}{2^{n+2}}\right)$$
The positive sign of the square root above, while appling $(1)$, is justified by $\frac{\pi}{4}>\frac{\pi}{4}-J_n\frac{\pi}{2^{n+2}}>0$, where the $\sin$ function is positive. As a result:
$$2\sin\left(J_{n}\frac{\pi}{2^{n+1}}\right)=\underbrace{\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}}_{n\text{ times}} \tag{2}$$

Jacobsthal sequence also has a closed form of 
$$ J_n = \frac{2^n - (-1)^n}{3}$$
which can be solved using, for example, characteristic polynomials (more than half of the work is done here), leading to
$$2\sin\left(\frac{2^n-(-1)^n}{2^{n+1}}\cdot\frac{\pi}{3}\right)=\underbrace{\sqrt{2-\sqrt{2-\sqrt{2-...-\sqrt{2-\sqrt{2}}}}}}_{n\text{ times}} \tag{3}$$

Remark: It is worth noting this question is not duplicating this family of questions.
