Question 124, Given the recurrence relation $a_{n+1}=\sqrt{\frac{a_{n}+2-\sqrt{2-a_{n}}}{2}}(n \in N), a_{0}=\frac{\sqrt{2}}{2}$, try to find $a_{n}$.


Question 124, Solution: Let $\mathrm{a}_{\mathrm{n}}=\sqrt{2} \cos \theta_{\mathrm{n}}$, then $\theta_{0}=\frac{\pi}{3}$. According to the conditions, we have:
$$
\begin{array}{l}
\sqrt{2} \cos \theta_{n+1}=\sqrt{\frac{\sqrt{2} \cos \theta_{n}+2-\sqrt{2-\left(\sqrt{2} \cos \theta_{n}\right)^{2}}}{2}}=\sqrt{\frac{\sqrt{2} \cos \theta_{n}+2-\sqrt{2} \sin \theta_{n}}{2}}=\sqrt{1+\cos \left(\theta_{n}+\frac{\pi}{4}\right)} \\
\Rightarrow \cos \theta_{n+1}=\cos \frac{\theta_{n}+\frac{\pi}{4}}{2} \\
\Rightarrow \theta_{n+1}=\frac{\theta_{n}+\frac{\pi}{2}}{2} \\
\Rightarrow 2\left(\theta_{n+1}-\frac{\pi}{4}\right)=\theta_{n}-\frac{\pi}{4} \\
\Rightarrow \theta_{n}=\frac{\pi}{4}+\frac{\frac{\pi}{3} 4}{2^{n}}=\frac{\pi}{4}+\frac{\pi}{3 \cdot 2^{n+2}}
\end{array}
$$

Thus, we know $a_{n}=\sqrt{2} \cos \left(\frac{\pi}{4}+\frac{\pi}{3 \cdot 2^{n+2}}\right)$.