# Simplifying $\sqrt{2+e^{8t}+e^{-8t}}$

I need to simplify this radical $$\sqrt{2+e^{8t}+e^{-8t}}$$

How is this done? I do not know where to go from here to simplify this further.

Hint: Use the identity $(x+x^{-1})^2 = x^2+2+x^{-2}$ for $x = e^{4t}$.
Let's set $$e^t$$ as $$T$$. Then it'll be $$\sqrt{T^8+2+\frac{1}{T^8}}$$. Using the factorization rule $$x^2+2+\frac{1}{x^2}=(x+\frac{1}{x})^2$$, it'll be $$e^{4t}+\frac{1}{e^{4t}}$$ assuming that $$t$$ is real.
$$\sqrt{4.\cosh(4t)^{2}}$$ $$=2.\cosh(4t)$$ $$=e^{4t}+e^{-4t}$$