# Simplifying a Hyperbolic function

• January 20th 2010, 07:46 AM
Beard
Simplifying a Hyperbolic function
I'm meant to put coth(x) in terms of e^x and e^-x. I keep end up getting the wrong answer and I have no idea how to do it.

My final answer is $\frac{e^{2x} + 2 + e^{-2x}}{e^{2x} - e^{-2x}}$

Those of you that can actually do this probably thinking 'How on earth did he come to that?'
• January 20th 2010, 08:55 AM
Drexel28
Quote:

Originally Posted by Beard
I'm meant to put coth(x) in terms of e^x and e^-x. I keep end up getting the wrong answer and I have no idea how to do it.

My final answer is $\frac{e^{2x} + 2 + e^{-2x}}{e^{2x} - e^{-2x}}$

Those of you that can actually do this probably thinking 'How on earth did he come to that?'

$\text{coth}(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{\fr ac{e^{x}-e^{-x}}{2}}{\frac{e^{x}+e^{-x}}{2}}=\frac{e^x -e^{-x}}{e^{x}+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}$
• January 20th 2010, 08:57 AM
Manx
Can you show what you did?
The answer looks close to me, but the form you used puzzles me.
• January 20th 2010, 08:58 AM
Drexel28
Quote:

Originally Posted by Manx
Can you show what you did?
The answer looks close to me, but the form you used puzzles me.

The only part that may be unclear is that I arrived at the last form by multiplying by $\frac{e^x}{e^x}$
• January 20th 2010, 12:28 PM
Beard
Quote:

Originally Posted by Drexel28
The only part that may be unclear is that I arrived at the last form by multiplying by $\frac{e^x}{e^x}$

That was where i was going wrong because I knew i had to multiply by 1 (effectively 1 anyway) but I was doing

$\frac{e^x + e^{-x}}{e^x + e^{-x}}$

Which you may think is ridiculous but the only other time i had used that method was in removing surds from the denominator and you changed the sign of what you were multiplying it by to remove the sqrt.

Anyway, thankyou for that and thanks to Manx for asking the exact same thing as I would have done lol.

(Clapping)