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))
Well, first you'd need to move the 3 to the other side and then take the inverse:
$\displaystyle x = arctanh\left(\frac{1}{3}\right)$
There is a formula for solving the arctanh, and that is:
$\displaystyle arctanh(u) = \frac{1}{2}ln\left(\frac{1 + u}{1 - u}\right)$
Plugging in u = 1/3, you get:
$\displaystyle x = arctanh\left(\frac{1}{3}\right) = \frac{1}{2}\left[ln\left(1 + \frac{1}{3}\right) - ln\left(1 - \frac{1}{3}\right)\right]$
$\displaystyle x = \frac{1}{2}\left[ln\left(\frac{4}{3}\right) - ln\left(\frac{2}{3}\right)\right]$
This ends up being:
$\displaystyle x \approx 0.347$
Plug this into sinh and cosh to get:
$\displaystyle sinh(0.347) \approx 0.354$
$\displaystyle \frac{1}{2\sqrt{2}} \approx 0.354$
That checks out with the answer.
$\displaystyle cosh(0.347) \approx 1.061$
$\displaystyle \frac{3}{2\sqrt{2}} \approx 1.061$
This also checks out.
Hope this helps!
Hello heatlyA more direct way is to use the relationships:
$\displaystyle \tanh x = \dfrac{\sinh x}{\cosh x}$ ...(1)and
$\displaystyle \cosh^2x-\sinh^2x = 1$ ...(2)So
$\displaystyle 3\tanh x=1$Then substitute into (2):
$\displaystyle \Rightarrow\tanh x=\dfrac{\sinh x}{\cosh x} = \frac13$, using (1)
$\displaystyle \Rightarrow \sinh x = \frac13\cosh h$
$\displaystyle \cosh^2x-\frac19\cosh^2x = 1$
Can you complete it from here?
Grandad