Another way to evaluate the nested radical $x=\sqrt{2+{\sqrt{2+\sqrt{2+\ldots}}}}$

Question

Consider the following expression: $$x=\sqrt{2+{\sqrt{2+\sqrt{2+\ldots}}}}$$ It can easily be evaluated by realizing that: $$x=\sqrt{2+x}$$ After squaring both sides and solving the quadratic equation, we find that $x=2$. But my question is: out of curiosity, is there any other way to evaluate this nested radical? Perhaps one involving infinite series/products?

Answer 1

Hint Prove by induction that

$$\sqrt{2+{\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}=2 \cos \left(\frac{\pi}{2^{n+1}}\right)$$

where $n$ is the number of roots.

P.S. If I am not mistaken, this can be interpreted geometrically as follows:

If $A_1A_2....A_{2^n}$ is a regular polynomial with $A_1A_2=1$ then $$A_1A_3=\sqrt{2+{\sqrt{2+\sqrt{2+\ldots\sqrt{2}}}}}$$

Now, as $n$ increases the angle $A_2$ gets closer and closer to $180$ and hence $A_1A_3$ gets closer and closer to $A_1A_2+A_2A_3$.