Question

Dumb Question: What is the inverse of the inverse of hyperbolic cosine?

Answer 1

Hmm it's probably just hyperbolic cosine, ya? 0_o

Answer 2

I am not sure anymore, so If I have something like this: What should I do?
\[7\times \cosh^{-1}(2) = \cosh^{-1}\left( \sqrt{\frac{10 ^{\frac{A _{s} }{ 10 }}-1 }{ 10^{\frac{A _{P} }{ 10 }}-1 }} \right) \]

Answer 3

I wan to move cosh-1 to where 7 is, will it be hyperbolic cosine?

Answer 4

@zepdrix I tried it and it works thanks

Answer 5

Applying cosh to each side,\[\large\rm \cosh(7\cosh^{-1}(2))\quad=\sqrt{\frac{10^{A_s/10}-1}{10^{A_p/10}-1}}\]

Wolfram is saying that:\[\large\rm \cosh(7\cosh^{-1}(2))\quad= 5042\]I'm trying to see how they got that though, hmm.

You can rewrite the inverse cosh in terms of natural logs in some way, yes?

Answer 6

Oh you got it? :O