Show that $ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$ Show that:
$$ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$$
My try:
As we can see that the LHS is in form of $\cos 8 \theta$ which must be converted into $\cos \theta $ in order to solve the equation.
There are several questions similar on this site but those aren't helping me much regarding this question.
So, I tried converting $\cos 8 \theta$ into $\cos \theta$ and then putting the value in the equation under square root and then further solving it into RHS.
So I got $\cos 8 \theta$ as:
$$ 2\cdot \{ 2\cdot [ 4 \cos ^4 \theta - 4.cos^2 \theta]^2 \}$$
So, I don't know how to solve it further.
Please help, putting the value of $\cos 8 \theta$ in place of that doesn't helps me much.
Thanks in Advance

Since $2(1+\cos x)=(2\cos x/2)^2$, $$\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}=\sqrt{2+\sqrt{2+2\cos 4\theta }}=\sqrt{2+2\cos 2\theta}=2\cos\theta,$$provided $\cos\theta,\,\cos 2\theta,\,\cos 4\theta>0$. In fact this implies $$\cos 2\theta =\sqrt{\frac{1+\cos 4\theta}{2}}\ge\frac{1}{\sqrt{2}},\,\cos\theta=\sqrt{\frac{1+\cos 2\theta}{2}}\ge\frac{\sqrt{2+\sqrt{2}}}{2}.$$
