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$\def\sl{\operatorname{sl}}\def\cl{\operatorname{cl}}\def\tl{\operatorname{tl}}\def\cscl{\operatorname{cscl}}\def\secl{\operatorname{secl}}\def\cotl{\operatorname{cotl}}\def\d{\,\mathrm{d}}$ The lemniscate sine "sl" and lemniscate cosine "cl" are defined as the solutions to:

$\displaystyle\frac{\partial}{\partial z}\,\sl(z)=\left(1+\sl^{2}(z)\right)\cl(z),\quad\frac{\partial}{\partial z}\,\cl(z)=-\left(1+\cl^{2}(z)\right)\sl(z),\quad \sl(0)=0,\quad \cl(0)=1$

or alternatively:

$\displaystyle\begin{aligned} \frac{\partial^2 y}{\partial z^2} &= -2y^3 \\ y &= \sl(z) = \operatorname{sn}(z\,\vert-1)\rightarrow y(0)=0, y^\prime(0)=1 \\ y &= \cl(z) = \operatorname{cd}(z\,\vert-1)\rightarrow y(0)=1, y^\prime(0)=0 \\ \end{aligned}$

where "sn" and "cd" are Jacobi elliptic integrals.

Looking at Wikipedia's article on the lemniscate functions, we see that:

$\begin{aligned} \sl(x+y) &= \frac{\sl(x)\cl(y)+\cl(x)\sl(y)}{1-\sl(x)\sl(y)\cl(x)\cl(y)} \\ \cl(x+y) &= \frac{\cl(x)\cl(y)-\sl(x)\sl(y)}{1+\sl(x)\sl(y)\cl(x)\cl(y)} \end{aligned}$

We know the formula for the circular tangent of the sum of two angles:

$\displaystyle\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$

If we define the "lemniscate tangent" as the ratio of the lemniscate sine to the lemniscate cosine in the same way the circular tangent is defined, i.e.

$\displaystyle\tl(z)=\frac{\sl(z)}{\cl(z)}=\operatorname{sn}(z\,\vert-1)\cdot\operatorname{dc}(z\,\vert-1)$

then this is what I've got for the lemniscate tangent's sum-of-angles identity:

$\begin{aligned} \tl(x+y) &= \frac{\sl(x+y)}{\cl(x+y)} \\ &= \frac{\frac{\sl(x)\cl(y)+\cl(x)\sl(y)}{1-\sl(x)\sl(y)\cl(x)\cl(y)}}{\frac{\cl(x)\cl(y)-\sl(x)\sl(y)}{1+\sl(x)\sl(y)\cl(x)\cl(y)}} \\ &= \underbrace{\left(\frac{\sl(x)\cl(y)+\cl(x)\sl(y)}{\cl(x)\cl(y)-\sl(x)\sl(y)}\right)}_{p}\underbrace{\left(\frac{1+\sl(x)\sl(y)\cl(x)\cl(y)}{1-\sl(x)\sl(y)\cl(x)\cl(y)}\right)}_{q} \end{aligned}$

I'm aware that you can divide the numerator and denominator of $p$ by $\cl(x)\cl(y)$, which gives us $\displaystyle p=\frac{\tl(x)+\tl(y)}{1-\tl(x)\tl(y)}$, reminiscent of the summation identity for the circular tangent.

My question is: Is it possible to express $\displaystyle q=\frac{1+\sl(x)\sl(y)\cl(x)\cl(y)}{1-\sl(x)\sl(y)\cl(x)\cl(y)}$ solely in terms of $\tl(x)$ and $\tl(y)$?

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  • $\begingroup$ The "Pythagorean-like identities" help. $\endgroup$ Commented yesterday
  • $\begingroup$ @Blue Could you tell me how this can be used to simplify $q$? $\endgroup$ Commented 4 hours ago

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Defining $s:=\operatorname{sl}(z)$, $c:=\operatorname{cl}(z)$, $t:=\operatorname{tl}(z)=\dfrac{s}{c}$, the Pythagorean-like relation $c^2+s^2+c^2s^2=1$ allows us to write $$s^2=\frac{1-c^2}{1+c^2} \quad\to\quad t^2=\frac{1-c^2}{c^2(1+c^2)}\quad\to\quad c=\pm\sqrt{\frac{-(1 + t^2) +\sqrt{(1+t^2)^2+4t^2}}{2 t^2}} \tag1$$ Then, we can note that $$sc=c^2t=\frac{-(1 + t^2) +\sqrt{(1+t^2)^2+4t^2}}{2 t} \tag2$$ This gives expressions to replace $\operatorname{sl}(x)\operatorname{cl}(x)$ and $\operatorname{sl}(y)\operatorname{cl}(y)$ in $q$.

The messy result simplifies with a bit rationalizing. Writing $t_x:=\operatorname{tl}(x)$ and $t_y:=\operatorname{tl}(y)$, we find $$\begin{align} q &= \frac{ t_x\sqrt{(1+t_y)^2 +4 t_y^2} + t_y\sqrt{(1+t_x)^2+4t_x^2}}{(t_x + t_y) (1 + t_x t_y)} \tag3 \\[8pt] &= \frac{ \sqrt{1+m_x^2} + \sqrt{1+m_y^2}}{m_x+m_y} \qquad \text{where}\quad m_z := \frac12\left(t_z+\dfrac1{t_z}\right) \tag4 \end{align}$$

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