Evaluate coshX given that tanhX Whilst working out some hyperbolic evaluation questions, I've come across this particular one. So far with any question I've come across I've simply tackled it step by step using hyperbolic identities. Before expressing my issue, here's the question.

Given that $\tanh x = \frac{2}{3}$, evaluate $\cosh x$, $\cosh 2x$ and $\tanh 2 x$
After seeing this I simply proceeded by trying to find out which is the correct identity to use to solve this question; however, I can't seem to find, or at least decide, which is the correct identity. My assumption is that I need to go through a series of identities to find out that matches my questions needs. The definition
$$\cosh x := \frac{ e^x + e^{-x}}{2}$$
was the first that came to mind, however I this clearly doesn't fit. The identity
$$\tanh x = \frac{\sinh x}{\cosh x}$$
doesn't get me anywhere either. Nor does
$$ \tanh^2 x + \text{sech}^2 x = 1 .$$
Any hint on how to tackle this would be of great help.
 A: Hint: use the second one, and the fact that $\cosh^2 - \sinh^2 = 1$.
A: The hint provided by Clement above leads you to the result easily, This is one alternative. You can also substitute the definition of tanhx i.e., in the exponential form and then solve for x. Now again to get the values of coshx and so on just substituting x value in their definition.
A: There are various identities, each of which has a trigonometric analogue, that will help in each situation:
For example, using the identity gives $1 - \tanh^2 x = \text{sech}^2 x$, so
$$\text{sech}^2 x = 1 - \left(\frac{2}{3}\right)^2 = \frac{5}{9},$$
and so
$$\color{red}{\cosh x} = \frac{1}{\text{sech} x} = \frac{1}{\frac{\sqrt{5}}{3}} \color{red}{= \frac{3}{\sqrt{5}}}.$$
Using this result and the "double angle" identities
$$\cosh 2 x = 2 \cosh^2 x - 1 \qquad \text{and} \qquad \tanh x = \frac{2 \tanh x}{1 + \tanh^2 x}$$
will give you the remaining quantities.
Interestingly, one can convert between the six elementary hyperbolic trigonometric functions by using a reference triangle as for the trigonometric case, treating the hyperbolic functions as their trigonometric analogues ($\cos$ for $\cosh$, etc.), subject to the one change the one labels the sides of the reference triangle with lengths so that the square of the label on the adjacent side, rather than the label on the hypotenuse, is equal to sum of the squares of the other two sides.
A: solving the equation $\tanh(x)=\frac{2}{3}$ we obtain
$x=artanh(\frac{2}{3})$
thus we obtain
$\cosh(artanh(\frac{2}{3}))=\frac{3}{\sqrt{5}}$
$\cosh(2artanh(\frac{2}{3}))=\frac{13}{5}$
$\tanh(2artanh(\frac{2}{3}))=\frac{12}{13}$
