9. In the sequence $\left\{a_{n}\right\}$, $a_{1}=\sqrt{2}, a_{n}=\sqrt{2+a_{n-1}}$, then $a_{n}=$
9. $2 \cos \frac{\pi}{2^{n+1}}$ By mathematical induction, it is not difficult to prove that $0<a_{n}<2(n=1,2, \cdots)$, so we can set $a_{n}=2 \cos \theta_{n}\left(0<\theta_{n}<\frac{\pi}{2}\right)$, thus $2 \cos \theta_{n}=\sqrt{2+2 \cos \theta_{n-1}}=2 \cos \frac{\theta_{n-1}}{2}$. Therefore, $\theta_{n}=\frac{1}{2} \theta_{n-1}$.
From $u_{1}=\sqrt{2}$, we get $\theta_{1}=\frac{\pi}{4}$.
Hence, $\theta_{n}=\theta_{1}\left(\frac{1}{2}\right)^{n-1}=\frac{\pi}{2^{n+1}}$. Therefore, $a_{n}=2 \cos \frac{\pi}{2^{n+1}}$.