# How is $\sqrt {2+\sqrt {2+\sqrt {2+}}} ... n$ times = $2\cos( π/2^{n+1})$?

How is $\sqrt {2+\sqrt {2+\sqrt {2+}}} ... n$ times = $2\cos( π/2^{n+1})$?

No idea. Please help. I found this identity in a solution of a problem related to limits. Also if any more identities like this then please let me know in the answer column.

• Did you try induction? It looks like the obvious first step...
– 5xum
Jan 28, 2016 at 14:36
• Yes, use induction and the half-angle formula for cosine. Jan 28, 2016 at 14:39
• You are missing $(1+cos\theta)$ in last bracket i suppose Jan 28, 2016 at 14:43
• Isn't this that beautiful result of Vieta ? Jan 28, 2016 at 15:20
• @5xum Actually I didn't want a proof by induction. Infact I wanted a synthetic/algebraic proof , that Leg has explained beautifully I suppose. Jan 28, 2016 at 16:56

Recall the identity $$2\cos(\theta) = \sqrt{2+2\cos(2\theta)}$$ This means $$2\cos(\theta) = \sqrt{2+\sqrt{2+2\cos(4\theta)}} = \sqrt{2+\sqrt{2+\sqrt{2+2\cos(8\theta)}}} \text{ and so on }$$ Setting $\theta = \dfrac{\pi}{2^{n+1}}$, and going on for $n$ terms, we see that we end up with $\cos(\pi/2)$, which is indeed zero.

EDIT All the square roots come with a positive sign, since each of the cosine term is of the form $\cos(\pi/2^k)$, where $k \geq 1$, i.e., $\pi/2^k \in [0,\pi/]$, where the cosine is non-negative.

• Also include a remark on why to choose the positive square-root. Jan 28, 2016 at 14:44

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

Keep on going...until you reach $n=2$.

one way is to use recurrence,

if $n=0$, then $0=0$... OK

suppose now that the relation, say $H_n$, is true and prove that $H_{n+1}$ is verified.

squaring the two sides of $H_{n+1}$

${(\sqrt{2+\sqrt{2+...}})}^2$ n+1 times = 2+$\sqrt{2+\sqrt{2+...}}$ n times

$= 2+2cos(\dfrac{\pi}{2^{n+1}})$

$= 4cos(\dfrac{\pi}{2^{n+2}})$ (according to Ron Gordon below in his hint)

...

Then $H_{n+1}$ is true and finally, you have thus your relation.