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.

  • 3
    $\begingroup$ Did you try induction? It looks like the obvious first step... $\endgroup$
    – 5xum
    Jan 28, 2016 at 14:36
  • 1
    $\begingroup$ Yes, use induction and the half-angle formula for cosine. $\endgroup$ Jan 28, 2016 at 14:39
  • $\begingroup$ You are missing $(1+cos\theta)$ in last bracket i suppose $\endgroup$ Jan 28, 2016 at 14:43
  • $\begingroup$ Isn't this that beautiful result of Vieta ? $\endgroup$
    – imranfat
    Jan 28, 2016 at 15:20
  • $\begingroup$ @5xum Actually I didn't want a proof by induction. Infact I wanted a synthetic/algebraic proof , that Leg has explained beautifully I suppose. $\endgroup$ Jan 28, 2016 at 16:56

3 Answers 3


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.

  • 1
    $\begingroup$ Also include a remark on why to choose the positive square-root. $\endgroup$
    – GEdgar
    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.


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