# Simplification of hyperbolic function

• Jan 10th 2011, 06:34 AM
Ant
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))

http://i1104.photobucket.com/albums/...ingproblem.jpg

Any help would be appreciated!
• Jan 10th 2011, 06:41 AM
Prove It
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 $\displaystyle \sinh{[2\,\textrm{arcsinh}\,{(\omega \tau)}]}$?
• Jan 10th 2011, 06:56 AM
Ant
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!