# 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?

## 3 Answers

It might be easiest to start with

\begin{align} \cosh^2(x/2)&=\left(\frac{e^{x/2}-e^{-x/2}}{2}\right)^2\\\\ &=\frac{e^x+e^{-x}+2}{4}\\\\ &=\frac{1+\cosh x}{2}\tag 1\\\\ \end{align}

Multiplying both sides of $(1)$ by $b^2$ and taking the square root reveals that

$$b\cosh(x/2)=\sqrt{\frac{b^2(1+\cosh x)}{2}} \tag 2$$

whereupon letting $a=b^2/2$ in $(2)$ leads to

$$\bbox[5px,border:2px solid #C0A000]{b\cosh(x/2)=\sqrt{a(1+\cosh x)}}$$

Setting $b=2$ in $(2)$ yields the coveted expression

$$\bbox[5px,border:2px solid #C0A000]{2\cosh(x/2)=\sqrt{2(1+\cosh x)}}$$

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.

$$2\cosh(x) + 2 = e^x + e^{-x} + 2 = e^{-x} \left( e^{2x} + 2e^{x} + 1 \right) = e^{-x} \left( e^x + 1 \right)^2 = \left( e^{-x/2} (e^x+1) \right)^2 = \left( e^{x/2} + e^{-x/2} \right)^2 = \left( 2\cosh(x/2) \right)^2$$