# Math Help - Simplification of hyperbolic function

1. ## Simplification of hyperbolic function

I'm having trouble seeing how this piece of manipulation follows:

I can see that a factor of a 1/2 has been taken out but otherwise I'm unsure how to deal with the sinh(2arcsinh(wr))

Any help would be appreciated!

2. It's because $\displaystyle \sinh{2x} = 2\sinh{x}\cosh{x}$. In this case, $\displaystyle x = \arcsin{(\omega \tau)}$.

BTW, are you sure that it's $\displaystyle \sinh{[2\arcsin{(\omega \tau)}]}$ and not $\displaystyle \sinh{[2\,\textrm{arcsinh}\,{(\omega \tau)}]}$?

3. I hadn't even noticed it said sin. I'm pretty much certain that's a typo and as you say, it's supposed to be arcsinh.

Thanks!