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

Question

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.

Answer 1

Hint: Use the identity $(x+x^{-1})^2 = x^2+2+x^{-2}$ for $x = e^{4t}$.

Answer 2

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.

Answer 3

Assuming knowledge of hyperbolic functions, the function in question is nothing more than

$$\sqrt{4.\cosh(4t)^{2}}$$ $$=2.\cosh(4t)$$ $$=e^{4t}+e^{-4t}$$