Determining $a$ such that adjacent loops in the parameterized curve $(x,y) = (t+a\sin2t, \sin t)$ are tangent

Question

Consider the parametric equations of the independent parameter $t$ as follows:

$$\begin{cases} x = t + a \sin(2t) \\ y = \sin(t)\end{cases}$$

• When $a = 0$, we get the usual sine curve.

• When $a = 0.4$ we get a near saw-tooth curve.

• When $a = 0.5$ the peaks of the curve become cusped

• When $a > 0.5$, loops form.

• When $a \approx 3.895$ the adjacent loops become tangent.

• For values of $a > 3.895$ (approx) the loops overlap.

Question: Is there a way to calculate this "tangent" value of $a \approx 3.895$ more precisely and analytically, or can it only be found by numerical means?

Answer 1

The points of tangency occur where the tangent to the curve is vertical, i.e. when

$$\begin{align*} \frac{dx}{dt} &= 1 + 2a \cos(2t) = 0 \\ \implies t &= \pm \frac12 \cos^{-1}\left(-\frac1{2a}\right) + n\pi \\ &= \pm \left(\frac12 \csc^{-1}(2a) + \frac\pi4\right) + n\pi \\ \end{align*}$$

Fix $n=0$ for simplicity, so one of these points (red) is attained when $t=\dfrac12\csc^{-1}(2a)+\dfrac\pi4$, and we attain the same point again (blue) in the next period of the curve when $t=-\dfrac12\csc^{-1}(2a)+\dfrac{11\pi}4$.

The $x$-coordinates of these points are identical, so we must have

$$x\left(\frac12\csc^{-1}(2a)+\frac\pi4\right) = x\left(-\frac12\csc^{-1}(2a)+\frac{11\pi}4\right) \\[2ex] \implies \csc^{-1}(2a) + a \sin\left(\frac\pi2+\csc^{-1}(2a)\right) = \frac{5\pi}2 + a \sin\left(\frac{11\pi}2-\csc^{-1}(2a)\right) \\ \implies \csc^{-1}(2a) + 2 a \cos\left(\csc^{-1}(2a)\right) = \frac{5\pi}2 \\ \implies \csc^{-1}(2a) + \sqrt{4a^2-1} = \frac{5\pi}2$$

Solving for $a$ exactly seems unlikely.

Answer 2

Starting from @user170231's answer, we are looking for the zero of function $$f(a)= \csc^{-1}(2a) + \sqrt{4a^2-1}- \frac{5\pi}2 $$ $$\csc\big(f(a)\big)=\frac{2 a}{\sqrt{4 a^2-1} \sin \left(\sqrt{4 a^2-1}\right)+\cos \left(\sqrt{4 a^2-1}\right)}$$ Multiply by the conjugate of the denominator to face the problem of solving for $x$ $$\sin(x)-x\,\cos(x)=0 \qquad \text{with}\qquad x=\sqrt{4a^2-1}$$

The $n^{\text{th}}$ root of $\tan(x)=x$ is well known; it write $$x_{(n)}=q-\frac 1q \sum_{k=0}^\infty \frac {\alpha_k}{\beta_k} \frac 1{q^{2k}}\qquad \text{with}\qquad q=(2n+1)\frac \pi 2$$ Coefficients $\alpha_k$ and $\beta_k$ form sequences $A079330$ and $A088989$ in $OEIS$.

Because of the $\frac {5\pi}2$, we need to use $n=2$ which means $q=\frac {5\pi}2$ and the solution is then $$a=\frac 12 \sqrt{1+x^2_{(2)}}$$

Working now the expansion of $a$, we have, the simple
$$a=\frac 12 \Bigg(q-\frac 1q \sum_{k=0}^\infty \frac {\alpha_k}{\beta_k} \frac 1{q^{2k}}\Bigg)\qquad \text{with}\qquad q=\frac {5\pi}2$$

Vaclav Kotesovec wrote that $$\frac {\alpha_k}{\beta_k} \propto \frac{\left(\frac{\pi }{2}\right)^{2 k}}{k^{4/3}}\qquad \implies \qquad \frac{\frac {\alpha_{k+1}}{\beta_{k+1}} \frac 1{q^{2(k+1)}}} {\frac {\alpha_{k}}{\beta_{k}} \frac 1{q^{2k}}}=\frac{1}{25} \left(\frac{n}{n+1}\right)^{4/3}$$

So, a fast convergence for the summation. Let $p$ be the level of expansion of $\tan(x)-x$. The results are $$\left( \begin{array}{cc} p & a_{(p)} \\ 1 & 3.89554998993 \\ 2 & 3.89503823412 \\ 3 & 3.89485935285 \\ 4 & 3.89485434524 \\ 5 & 3.89485296250 \\ 6 & 3.89485289919 \\ 7 & 3.89485288482 \\ 8 & 3.89485288394 \\ 9 & 3.89485288376 \\ 10 & 3.89485288375 \\ \end{array} \right)$$