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
 A: 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 $$\cos 2\theta =\sqrt{\frac{1+\cos 4\theta}{2}}\ge\frac{1}{\sqrt{2}},\,\cos\theta=\sqrt{\frac{1+\cos 2\theta}{2}}\ge\frac{\sqrt{2+\sqrt{2}}}{2}.$$
A: Hint: use $$1+\cos 2x = 2\cos^2x$$
A: $$\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}$$
Firstly, $2+2cos8\theta=2+2(2cos^24\theta-1)=4cos^24\theta$
$$\therefore\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}=\sqrt{2+\sqrt{2+2cos4\theta}}$$
Then, $2+2cos4\theta=2+2(2cos^22\theta-1)=4cos^22\theta$
$$\therefore\sqrt{2+\sqrt{2+2cos4\theta}}=\sqrt{2+2cos2\theta}$$
Finally, $2+2cos2\theta=2+2(2cos^2\theta-1)=4cos^2\theta$
$$\therefore\sqrt{2+\sqrt{2+2cos2\theta}}=\sqrt{4cos^2\theta}=2cos\theta$$
A: You should notice that this is a (three times) repeated application of
$$\sqrt{2+2\cos 2\theta}=2\cos\theta$$ which is established by
$$\sqrt{2+2\cos2\theta}=\sqrt{2+4\cos^2\theta-2}.$$
A: It's interesting to note that the equality holds for only a limited range of values as pointed out in J.G.'s answer. See graph below. 

