Show that sequence $\sqrt{2+\sqrt{2+...+\sqrt{2}}}$ is equal to $2 \, \cos(\theta)$

Question

Basicly the title. The profesor gave me the indication $1+\cos(a) = 2 \, \cos²\left(\frac{a}{2}\right)$

I have to verify that the sequence can be written as that trigonometric fonction. Teta is between [0,pi/2]

Answer 1

Hint $$1+\cos(a) = 2 \, \cos²\left(\frac{a}{2}\right) \Rightarrow \\ 2+2\cos(a) = 4 \, \cos²\left(\frac{a}{2}\right) \Rightarrow \\ \sqrt{2+ 2\cos(a)} =2 \cos(\frac{a}{2})$$

This means that if you start with $$\sqrt{2}=2 \cos(\theta_1)$$ by induction you get $$\sqrt{2+\sqrt{2+...+\sqrt{2}}}=\sqrt{2+2\cos(\frac{\theta_1}{2^{n-1}})}=2\cos(\frac{\theta_1}{2^n})$$ where there are $n$ roots on the LHS.

What is $\theta_1$?

Answer 2

$\sqrt{2 + \sqrt{2 + ...}} = 2\cos\theta$

$\sqrt{2 + \sqrt{2 + ...}}^2 = 4\cos^2\theta$

$2 + \sqrt{2 + ...} = 4\cos^2\theta$

$\sqrt{2 + \sqrt{2 + ...}} = 4\cos^2\theta -2$

$2cos \theta = 4\cos^2\theta - 2$

$4 \cos^2 \theta - 2\cos \theta - 2 = 0$

$2\cos^2 \theta - \cos \theta - 1 = 0$

$(2\cos \theta + 1) (\cos \theta - 1) = 0$

$\cos \theta = \{1, -1/2\}$

$theta = \{0, 2\pi/3\}$