# tanh(x)

• Dec 2nd 2008, 09:40 AM
Greengoblin
tanh(x)
I'm trying to prove:

$\tanh(2x)=\frac{2\tanh(x)}{1+\tanh^2(x)}$

Here's what I've done so far:

$\tanh(2x)=\frac{\sinh(2x)}{\cosh(2x)}=\frac{\left( \frac{e^{2x}-e^{-2x}}{2}\right)}{\left(\frac{e^{2x}+e^{-2x}}{2}\right)}$

$=\frac{e^{2x}-e^{-2x}}{2}\frac{2}{e^{2x}+e^{-2x}}=\frac{(e^{2x}-e^{-2x})(e^{2x}+e^{-2x})+4}{2(e^{2x}+e^{-2x})}$

$=\frac{e^{4x}-e^{-4x}+4}{2(e^{2x}+e^{-2x})}$

From this point I'm not sure what to do. I know of the identities:

$\cosh(x)-\sinh(x)=e^{-x}$

$\cosh^2(x)+\sinh^2(x)=\cosh(2x)$

However, even with these in mind, I've been trying this for ages to no avail. Can someone give me some pointers?

Thanks
• Dec 2nd 2008, 10:20 AM
Moo
Hello,

Quote:

$\frac{e^{2x}-e^{-2x}}{2}\frac{2}{e^{2x}+e^{-2x}}=\frac{(e^{2x}-e^{-2x})(e^{2x}+e^{-2x})+4}{2(e^{2x}+e^{-2x})}
$

But this is false !

It's a multiplication, not an addition.

$\frac{e^{2x}-e^{-2x}}{2}+\frac{2}{e^{2x}+e^{-2x}}$ will give you what you wrote, but the multiplication will give :

(you can simplify by 2) :

$=\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$

But I'd suggest you use these formulae :
$\sinh(2x)=2 \cosh(x) \sinh(x)$
$\cosh(2x)=\cosh^2(x)+\sinh^2(x)$

Hence $\tanh(2x)=\frac{2 \cosh(x) \sinh(x)}{\cosh^2(x)+\sinh^2(x)}$

Divide both numerator and denominator by $\cosh^2(x)$ (Wink)
• Dec 2nd 2008, 10:39 AM
Greengoblin
Ahh, such as stupid mistake, I've been looking at it so long I missed it. (Giggle)

Thanks alot!
• Dec 2nd 2008, 11:32 AM
Greengoblin
Ok, now I'm trying to prove:

$\frac{(e^x+e^{-x})^2}{4(e^x-e^{-x})}=\sinh(x)$

Any ideas? I'm not sure if this is even right, it was after an integration of $\cosh(x)$
• Dec 2nd 2008, 12:54 PM
Soroban
Hello, Greengoblin!

This is not true . . .

$\frac{(e^x+e^{-x})^2}{4(e^x-e^{-x})}\;=\;
\frac{(e^x + e^{-x})^2}{4}\cdot\frac{1}{e^x-e^{-x}}$
. $= \;\left(\frac{e^x + e^{-x}}{2}\right)^2 \cdot\frac{1}{2}\cdot\frac{2}{e^x-e^{-x}} \;= \;\frac{1}{2}\,\frac{\cosh^2\!x}{\sinh x}$

• Dec 2nd 2008, 03:23 PM
Greengoblin
Hi Soroban, I suspected as much, since I was getting that same expression in trying to simplify. It must be a problem with my integration then. I used the substitution $u=e^x+e^{-x}$, and I think my mistake was taking the $\frac{1}{e^x+e^{-x}}$ out of the integral, which can't be done.

Is there some easier substitution I can make?
• Dec 2nd 2008, 03:55 PM
Greengoblin
Hi, no worries, I've cracked it now...proved that the derivative of sinhx is coshx, therefore the integral of coshx is sinhx. Thanks for the help though!