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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.

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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.

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  • $\begingroup$ Very nice post Travis, really helped out! $\endgroup$ – Juxhin Sep 26 '14 at 15:57
  • $\begingroup$ You're welcome, I'm glad it was useful. Wikipedia has a good list of basic hyperbolic identities: en.wikipedia.org/wiki/Hyperbolic_function As a general, rule, they mimic the more familiar trigonometric identities but with differing signs in places. $\endgroup$ – Travis Willse Sep 26 '14 at 16:39
  • $\begingroup$ Good point. Thanks again Travis $\endgroup$ – Juxhin Sep 27 '14 at 9:04
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Hint: use the second one, and the fact that $\cosh^2 - \sinh^2 = 1$.

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  • $\begingroup$ What is the relation though? I don't have any information given apart from the value of $tanh x = 2/3$ $\endgroup$ – Juxhin Sep 26 '14 at 15:29
  • $\begingroup$ You have $\sinh^2 = \cosh^2 - 1$ and thus $\tanh^2=\frac{\sinh^2}{\cosh^2} = 1-\frac{1}{\cosh^2}$. Can you continue? $\endgroup$ – Clement C. Sep 26 '14 at 15:32
  • $\begingroup$ Oh that's an interesting way to tackle it. Makes sense, will try working the rest out. Thanks Clement :) $\endgroup$ – Juxhin Sep 26 '14 at 15:34
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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.

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solving the equation $\tanh(x)=\frac{2}{3}$ we obtain $x=artanh(\frac{2}{3})$ thus we obtain $\cosh(artanh(\frac{2}{3}))=\frac{3}{\sqrt{5}}$
$\cosh(2artanh(\frac{2}{3}))=\frac{13}{5}$
$\tanh(2artanh(\frac{2}{3}))=\frac{12}{13}$

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  • $\begingroup$ Oh this is inverse hyperbolics if I'm not mistaken. However I am to do it with the regular hyperbolic functions $\endgroup$ – Juxhin Sep 26 '14 at 15:33

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