anonymous
  • anonymous
Does anyone know where you could find the deduction of the inverse of tanh(x) starting from the standard hyperbolic trigonometry definitions? (Or does anyone possibly know how to deduce it from them?)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
@hartnn , @TuringTest , @AravindG Please help ;_;
anonymous
  • anonymous
I basically get it, down to tanh(x)=(e^(2x)-1)/(e^(2x)+1)
AravindG
  • AravindG
It is right isnt it?

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anonymous
  • anonymous
Indeed, but how do you go from here to the inverse of tanh(x)?
hartnn
  • hartnn
do u know componendo - dividendo ?
AravindG
  • AravindG
Take tanh x=y. Then apply log on both sides.
AravindG
  • AravindG
Solve for x.
hartnn
  • hartnn
if a/b = c/d then (a+b) /(a-b) = (c+d)/ (c-d) did u know this ?
anonymous
  • anonymous
That's what I did, I didn't get anywhere on the log expression though. Hmm, not sure hartnn.
hartnn
  • hartnn
that rule is called componendo-dividendo apply that for y/1= (e^(2x)-1)/(e^(2x)+1)
anonymous
  • anonymous
I'd have to give a proof in order to utilize it though. D: It's supposed to be solved without it.
hartnn
  • hartnn
that will isolate e^2x and hence you can apply log
hartnn
  • hartnn
solved without what ?
anonymous
  • anonymous
No shortcuts. :D
anonymous
  • anonymous
Algebraically
AravindG
  • AravindG
Great! Good work!
hartnn
  • hartnn
lol ok, don't apply that rule (bdw, that was algebraic property) y= (e^(2x)-1)/(e^(2x)+1) y (e^2x +1) = e^2x -1 y e^2x + y = e^2x -1 y e^2x -e^2x = -1-y need i go further ?
hartnn
  • hartnn
hint for next step factor out e^2x from on left side
hartnn
  • hartnn
but did u get what i did ?
hartnn
  • hartnn
\(\huge \dfrac{1}{2}\ln (\dfrac{1+x}{1-x})\) you should get this
hartnn
  • hartnn
-1-y / y-1 = (1+y)/ (1-y) the 1/2 will come outside ln right ? because 2x= ln (...)
anonymous
  • anonymous
What happens after you apply ln?
hartnn
  • hartnn
e^2x =[(1+y)/(1-y) ] clear till here ??
anonymous
  • anonymous
Man I feel stupid. D:
anonymous
  • anonymous
Absolutely.
hartnn
  • hartnn
any more doubts?
anonymous
  • anonymous
Nope, thanks a lot! :)
hartnn
  • hartnn
welcome ^_^

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