Evaluate the following $\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}$

To evaluate the given expression, we will simplify it step-by-step from the innermost square root to the outermost square root using the half-angle trigonometric identity:
$1+\cos 2A=2{\cos }^{2}A$
which can also be written as:
$2+2\cos 2A=4{\cos }^{2}A$

Step 1: Simplify the innermost radical expression
Consider the innermost term under the first square root:
$2+2\cos 8\theta$
Using the identity with $2A=8\theta$ (so $A=4\theta$), we get:
$2+2\cos 8\theta =2(1+\cos 8\theta )=2\left(2{\cos }^{2}4\theta \right)=4{\cos }^{2}4\theta$
Taking the square root of this term:
$\sqrt{2+2\cos 8\theta }=\sqrt{4{\cos }^{2}4\theta }=2\cos 4\theta$

Step 2: Simplify the middle radical expression
Substitute the result from Step 1 back into the expression:
$\sqrt{2+\sqrt{2+2\cos 8\theta }}=\sqrt{2+2\cos 4\theta }$
Using the identity again with $2A=4\theta$ (so $A=2\theta$):
$2+2\cos 4\theta =4{\cos }^{2}2\theta$
Taking the square root:
$\sqrt{2+2\cos 4\theta }=\sqrt{4{\cos }^{2}2\theta }=2\cos 2\theta$

Step 3: Simplify the outermost radical expression
Now substitute the result from Step 2 into the outer square root of the original expression:
$\sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}=\sqrt{2+2\cos 2\theta }$
Applying the identity one final time with $2A=2\theta$ (so $A=\theta$):
$2+2\cos 2\theta =4{\cos }^2\theta$
Taking the square root:
$\sqrt{2+2\cos 2\theta }=\sqrt{4{\cos }^2\theta }=2\cos \theta$

Therefore, the simplified value of the given expression is:
$2\cos \theta$