Help with a hyperbolic trig problem

$$\tanh n=\operatorname{csch}n$$ 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.

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

$$\iff\frac{e^{2n}-1}{e^{2n}+1}=\frac{2e^n}{e^{2n}-1}$$

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 $a^2,$ $$a^2-2a-2-\frac2a+\frac1{a^2}=0$$

$$\iff\left(a^2+\frac1{a^2}\right)-2\left(a+\frac1a\right)-2=0$$

$$\iff\left(a+\frac1a\right)^2-2-2\left(a+\frac1a\right)-2=0$$

Hope you can take it from here?

• ya, that's perfect thanks – user183782 Oct 26 '14 at 12:23
• it is $a^4-2a^3-2a^2-2a+1=0$ – Dr. Sonnhard Graubner Oct 26 '14 at 12:36
• @Dr.SonnhardGraubner, Thanks for your observation – lab bhattacharjee Oct 26 '14 at 12:41
• 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? – user183782 Oct 26 '14 at 13:23
• @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 $x?$ – 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 $a^4-2a^3-2a^2-2a+1=0$

• I was trying to get it to quadratic form. – 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:

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

so

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

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:

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