1. ## tanh(x)

I'm trying to prove:

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

Here's what I've done so far:

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

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

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

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

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

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

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

Thanks

2. Hello,

$\displaystyle \frac{e^{2x}-e^{-2x}}{2}\frac{2}{e^{2x}+e^{-2x}}=\frac{(e^{2x}-e^{-2x})(e^{2x}+e^{-2x})+4}{2(e^{2x}+e^{-2x})}$
But this is false !

It's a multiplication, not an addition.

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

(you can simplify by 2) :

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

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

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

Divide both numerator and denominator by $\displaystyle \cosh^2(x)$

3. Ahh, such as stupid mistake, I've been looking at it so long I missed it.

Thanks alot!

4. Ok, now I'm trying to prove:

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

Any ideas? I'm not sure if this is even right, it was after an integration of $\displaystyle \cosh(x)$

5. Hello, Greengoblin!

This is not true . . .

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

6. 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 $\displaystyle u=e^x+e^{-x}$, and I think my mistake was taking the $\displaystyle \frac{1}{e^x+e^{-x}}$ out of the integral, which can't be done.

Is there some easier substitution I can make?

7. 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!