Simplification of $\sqrt{2\cosh(x)+2}$ I do not understand the simplification 
$$
\sqrt{2\cosh(x)+2}=2\cosh(x/2)
$$
More generally, I do not understand why 
$ \sqrt{a\cosh(x)+a}=b\cosh(x/2)$
What is the relationship between $a$ and $b$?
How does adding $a$ remove the radical?

We have that: $$\cosh(2x) = \cosh^2x + \sinh^2x = 2\cosh^2x - 1,$$since $\cosh^2x - \sinh^2x=1$. In that identity, we make $x \mapsto x/2$, so we obtain: $$\cosh x = 2 \cosh^2\left(\frac x 2\right)-1 \implies \cosh\frac{x}{2} = \sqrt{\frac{1+\cosh x}{2}}.$$Multiplying both sides by $2$ we have: $$2 \cosh \frac x 2 = 2\sqrt{\frac{1+\cosh x}{2}} = \sqrt{\frac{4(1+\cosh x)}{2}} = \sqrt{2+2\cosh x}.$$
The second expression there does not makes sense. What you can do is copy the above work multiplying by $\sqrt{2a}$, if $a>0$, to obtain: $$ \sqrt{2a}\cosh \frac x 2 = \sqrt{a+a\cosh x},$$in the same fashion.
