# Simplifying $\frac{\cosh(4n\theta)-1}{\sinh((2n+1)\theta)+\sinh((2n-1)\theta)}$

I've come across this expression as part of a very separate problem I've been working on: $$\frac{\cosh(4n\theta)-1}{\sinh((2n+1)\theta)+\sinh((2n-1)\theta)}$$ and noticed that whenever one sets $$\theta: \sinh(\theta)\in \Bbb N$$, the expression always comes out an integer. This leads me to believe there is a nice(-ish?) simplification for this, but I've had no luck finding it thus far.

I know the denominator can be rewritten as $$2\sinh(2n\theta)\cosh(\theta)$$ but I couldn't see this getting anywhere.

• Maybe you can use exponentials Nov 13, 2022 at 17:59
• Using formulae for multiple arguments of the hyperbolic functions, your expression is just $\sinh(2n \theta) / \cosh(\theta)$ Nov 13, 2022 at 18:04

$$\cosh 2x + 1 = \frac {e^{2x} + e^{-2x} - 2}{2} = \frac {(e^{x} - e^{-x})^2}{2} = 2\sinh^2 x$$

The numerator can be rewritten as $$2\sinh^2 2n\theta$$

As for the denominator

$$\cosh a \sinh b = \frac 14 (e^{a}+e^{-a})(e^{b}-e^{-b}) = \frac 14(e^{a+b}-e^{a-b}+e^{b-a}-e^{-(a+b)}) = \frac 12(\sinh(a+b)-\sinh(a-b)) = 2(\sinh(a+b)+\sinh(b-a)$$

$$a+b = 2n\theta+\theta\\ -a+b = 2n\theta - \theta\\ a = \theta\\ b = 2n\theta$$

$$\sin(2n\theta +1) + \sin(2n\theta - 1) = 2\cosh \theta\sinh 2n\theta$$

$$\frac {2\sinh^2 2n\theta}{2(\cosh\theta\sinh 2n\theta)} = \frac {\sinh 2n\theta}{\cosh\theta}$$

Well, this is just

$$\text{sech}(\theta ) \sinh (2 \theta n)$$

Let $$u=e^{\theta}$$. If $$\sinh\theta=\frac{u-u^{-1}}{2}$$ is an integer then $$u-u^{-1}$$ is an (even) integer. By successive squaring we find that $$u^{2k}+u^{-2k}$$ is integer for all $$k\in\Bbb{N}$$.

The expression is equal to $$sech\theta\sinh(2n\theta)=\frac{u^{2n}-u^{-2n}}{u+u^{-1}}=\frac{(u^2)^{n}-(u^{-2})^{n}}{u+u^{-1}}=\frac{(u^{2}-u^{-2})\left(\sum_{k=0}^nu^{2n-2k}u^{-2k}\right)}{u+u^{-1}}=(u-u^{-1})\left(\sum_{k=0}^nu^{2n-4k}\right)=(u-u^{-1})\left(\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}(u^{2n-4k}+u^{-(2n-4k)})+\frac{(-1)^n+1}{2}\right)$$ which is an integer.