Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct?

Looks great. Here is a fun trick I've seen to answer this question.

Using the half angle formula, notice the following:

$$\cos\left(\frac{\pi}{4}\right)=\frac{1}{2}\sqrt 2\\\cos\left(\frac{\pi}{8}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt 2)}=\frac{1}{2}\sqrt{2+\sqrt 2}\\\cos\left(\frac{\pi}{16}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt{ 2+\sqrt2})}=\frac{1}{2}\sqrt{2+\sqrt {2+\sqrt 2}}\\\vdots\\\cos\left(\frac{\pi}{2^{n+1}}\right)=\underbrace{\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+{\ldots}}}}}_\text{n times}=\frac{1}{2}a_{n}$$

Now let $n$ approach infinity.

A somewhat shorter solution: prove by induction that $$a_n=2\cos \left(2^{-1-n}\pi\right).\tag{1}$$ Then, obviously, $\displaystyle \lim_{n\rightarrow\infty}a_n=2$.

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Added: To understand the origin of (1), substitute $a_n=2\cos\alpha_n$ into the recursion relation.