1. ## Hyperbolic functions

If 3tanhx = 1 ,determine exact values of sinhx and coshx.

Can anybody help with this one!

The ans is,

1/(2(sqrt 2)) and 3/(2(sqrt2))

2. Well, first you'd need to move the 3 to the other side and then take the inverse:

$x = arctanh\left(\frac{1}{3}\right)$

There is a formula for solving the arctanh, and that is:

$arctanh(u) = \frac{1}{2}ln\left(\frac{1 + u}{1 - u}\right)$

Plugging in u = 1/3, you get:

$x = arctanh\left(\frac{1}{3}\right) = \frac{1}{2}\left[ln\left(1 + \frac{1}{3}\right) - ln\left(1 - \frac{1}{3}\right)\right]$

$x = \frac{1}{2}\left[ln\left(\frac{4}{3}\right) - ln\left(\frac{2}{3}\right)\right]$

This ends up being:

$x \approx 0.347$

Plug this into sinh and cosh to get:

$sinh(0.347) \approx 0.354$

$\frac{1}{2\sqrt{2}} \approx 0.354$

That checks out with the answer.

$cosh(0.347) \approx 1.061$

$\frac{3}{2\sqrt{2}} \approx 1.061$

This also checks out.

Hope this helps!

3. Hello heatly
Originally Posted by heatly
If 3tanhx = 1 ,determine exact values of sinhx and coshx.

Can anybody help with this one!

The ans is,

1/(2(sqrt 2)) and 3/(2(sqrt2))
A more direct way is to use the relationships:
$\tanh x = \dfrac{\sinh x}{\cosh x}$ ...(1)
and
$\cosh^2x-\sinh^2x = 1$ ...(2)
So
$3\tanh x=1$

$\Rightarrow\tanh x=\dfrac{\sinh x}{\cosh x} = \frac13$, using (1)

$\Rightarrow \sinh x = \frac13\cosh h$
Then substitute into (2):
$\cosh^2x-\frac19\cosh^2x = 1$

Can you complete it from here?