<s>[INST] 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
 [/INST] $$\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}$$
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, $2+2cos2\theta=2+2(2cos^2\theta-1)=4cos^2\theta$
$$\therefore\sqrt{2+\sqrt{2+2cos2\theta}}=\sqrt{4cos^2\theta}=2cos\theta$$
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