# Calculation with Hyperbolic Cosine

Could you please check my work?

$\cosh \left(\ln \sqrt{5}\right) =\ ?$

\begin{align*}\cosh(x) &= \frac{e^x + e^{-x}}{2} \\ \\ \frac{e^{\ln \sqrt{5}} + e^{-\ln \sqrt{5}}}{2} &= \frac{\sqrt{5} + \frac{1}{\sqrt{5}}}{2}\\ &=\frac{\sqrt{5}}{2}+\frac{1}{2\sqrt{5}} \\ &= \frac{3}{\sqrt{5}} \end{align*}

Thanks.

• $\,\checkmark$ – David Dec 4 '15 at 0:49
• As you can see, not using Mathjax has caused quite a bit of confusion among people answering your question. Please take the time to learn. – pjs36 Dec 4 '15 at 2:10

Umm...this seems like it has an error to me, unless I misread something. (There appear to be some LaTeX problems, so it's entirely possible I've misread.)

But

\begin{align} \cosh \left( \ln \sqrt{5} \right) & = \frac{1}{2} \left( e^{\ln \sqrt{5}}+e^{-\ln \sqrt{5}} \right) \\ & = \frac{1}{2} \left( \sqrt{5} + \frac{1}{\sqrt{5}} \right) \\ & = \frac{1}{2} \left( \frac{5+1}{\sqrt{5}} \right) \\ & = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \end{align}

• Yeah, it was originally written in plaintext, and every edit since has ... really not paid attention to how $\cosh$ is defined. Quite puzzling. – pjs36 Dec 4 '15 at 1:44

That's correct, no errors.

As a note, its well worth learning how to present questions well (like using the LaTex capabilities of this website, as all the editors have been doing). It makes it much easier for viewers to read and understand your question. It also encourages you to think about what's important in your question, as this determines a lot about how you present it.

• There are plenty errors! – Unit Dec 4 '15 at 1:27
• @Unit Most if not all were introduced by someone other than the OP editing the post. – pjs36 Dec 4 '15 at 1:46
• @pjs36 That's not what I see! math.stackexchange.com/revisions/1558985/1 – Unit Dec 4 '15 at 1:49