$$\tanh n=\operatorname{csch}n$$ Solve so that $n=\ln(x\pm x^{1/2})$ $%replace "x^{1/2}" with "\sqrt{x}" if you want. - editor$

I need some advice with this problem; I answered a similar one correctly but I can't get this one right. Here's my work so far:

\begin{align} \frac{\sinh n}{\cosh n}&=\frac1{\sinh n}\\ \frac{e^n-e^{-n}}{e^n+e^{-n}}&=\frac2{e^n-e^{-n}} \end{align}

Cross multiplying and simplifying got me to $$e^{2n}+e^{-2n}-2-2e^n-2e^{-n}=0$$

I know I need to get this to quadratic form, so I can set up the quadratic equation to get the answer, but I'm unsure how to do this. I'm not sure what to factor out to make it quadratic.

Some advice would be appreciated.


We have $$\frac{e^n-e^{-n}}{e^n+e^{-n}}=\frac2{e^n-e^{-n}}$$


Writing $e^n=a,$

$$\iff(a^2-1)^2=2a(a^2+1)\iff a^4-2a^3-2a^2-2a+1=0$$

As $a\ne0,$ like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$ divide either sides by $a^2,$ $$a^2-2a-2-\frac2a+\frac1{a^2}=0$$



Hope you can take it from here?

  • $\begingroup$ ya, that's perfect thanks $\endgroup$ – user183782 Oct 26 '14 at 12:23
  • $\begingroup$ it is $a^4-2a^3-2a^2-2a+1=0$ $\endgroup$ – Dr. Sonnhard Graubner Oct 26 '14 at 12:36
  • $\begingroup$ @Dr.SonnhardGraubner, Thanks for your observation $\endgroup$ – lab bhattacharjee Oct 26 '14 at 12:41
  • $\begingroup$ I'm stuck again, I got past the quadratic equation but I'm stuck on trying to take the ln of both sides. Any advice? $\endgroup$ – user183782 Oct 26 '14 at 13:23
  • $\begingroup$ @user183782, $$a+\frac1a=\pm\sqrt5+1$$ But for real $n,a=e^n>0\implies a+\frac1a\ge2$ $$\implies a+\frac1a=\sqrt5+1$$ Solve for $a$. But, what is $x?$ $\endgroup$ – lab bhattacharjee Oct 26 '14 at 13:29

you have $\frac{-e^{-x}+e^{x}}{e^{-x}+e^{x}}=\frac{2}{-e^{-x}+e^{x}}$ setting $e^{x}=a$ we get $\frac{-1/a+a}{1/a+a}=\frac{2}{-1/a+a}$ i will post my solution in a few minutes, simplifying this we obatin $a^4-2a^3-2a^2-2a+1=0$

  • $\begingroup$ I was trying to get it to quadratic form. $\endgroup$ – user183782 Oct 26 '14 at 12:52

I used a different approach that avoids the quartic. But it does involve complex numbers. I personally find it easier to view hyperbolic trig functions as circular trig functions of complex/imaginary arguments.

Using $\sin(ix) = i\sinh(x), \cos(ix) = \cosh(x)$, we can rewrite the original equation as:

$\displaystyle \frac{-i\sin(in)}{\cos(in)} = \frac{1}{-i\sin{in}}$


$\sin^2(in) + \cos(in) = 0$

Putting $in = m$ and using an elementary trig identity,

$\cos^2 m - \cos m - 1 = 0$

Solving for $\cos m$,

$\cos m = \frac{1}{2}(1 \pm \sqrt 5)$

Now reintroduce the hyperbolic trig function using $\cos m = \cos (in) = \cosh n$,

$\cosh n = \frac{1}{2}(1 \pm \sqrt 5)$

To simplify things, represent the surd on the RHS as $p$.

You have $\frac{1}{2}(e^n + e^{-n}) = p$.

Solving the quadratic in $e^n$ gets you:

$e^n = p \pm \sqrt{p^2-1}$

If you substitute the surd for $p$, you will find that $p^2-1 = p$, which is a pleasant "surprise" (not that it should be surprising if you notice the golden ratio form).

So $e^n = p \pm \sqrt p$

Giving $n = \ln(p \pm \sqrt p) = \ln(\frac{1}{2}(1 \pm \sqrt 5) \pm \sqrt{\frac{1}{2}(1 \pm \sqrt 5)})$

At this point, for real $n$, discard the inappropriate value of $p$, giving a final answer of:

$n = \ln(\frac{1}{2}(1 + \sqrt 5) \pm \sqrt{\frac{1}{2}(1 + \sqrt 5)})$


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