# Arctanh and arccoth

1. Jul 23, 2010

### Alexx1

How can I prove this:

1. For all y ∈ ]-1;1[ : Arctanh y = (1/2) ln( 1+y / 1-y )

2. For all y ∈ ]-∞;-1[ U ]1;+∞[ : Arccoth y = (1/2) ln( y+1 / y-1 )

Can I solve it by using this:

Arcsinh y = ln (x+square(x^2 +1))
Arccosh y = ln (x+square (x^2 -1))

2. Jul 23, 2010

### Office_Shredder

Staff Emeritus
Arctanhy is defined to be the number x such that tanh(x)=y. So to prove that arcanh(y)=(1/2) ln( 1+y / 1-y ), it's the same thing as showing that tanh((1/2) ln( 1+y / 1-y ))=y

3. Jul 23, 2010

### Alexx1

Thx, but I still don't know how I can prove that..

4. Jul 23, 2010

### Office_Shredder

Staff Emeritus
The definition of tanh is in terms of sinh and cosh, which are in terms of exponentials. Surely you can calculate an exponential raised to a logarithmic power, then there's just a bunch of algebra to do. If you get stuck post how far you've gotten and we can see how to progress

5. Jul 23, 2010

### Staff: Mentor

6. Jul 23, 2010

### Alexx1

I get :

2 tanh (1/2 ln (1+y / 1-y) = ... = (1+y)/2 - (1-y)/2

7. Jul 25, 2010

### Alexx1

I found it! Thank you very much!