A Sine Wave Where Alternate Distances Between 'Wave-center' Points Are Powers of φ This may be hard to visualize without my graph, see here
If $\phi=\left(\frac{1+5^{1/2}}{2}\right), \alpha=\phi^{-2}, \beta=1$, then the parametric equations, 
$$
(x, y)=\left(\sin(t)\cdot\left(\alpha\cdot\phi^{t-{\pi/2}/\pi}+\beta-\frac{\alpha}{\phi^{1/2}}\right)^{-1},\; \left(\alpha\cdot\phi^{t-{\pi/2}/\pi}+\beta-\frac{\alpha}{\phi^{1/2}}\right)\right)
$$ produce a graph where the vertical distances between points of tangency with $x·y=±1$ on alternate sides are powers of $\phi$. (when only positive numbers are graphed, starting at $\phi^{-1}$ and proceeding as follows: $\phi^{-1}, \phi^{0}, \phi^{1}, \phi^{2}, \phi^{3}$).
To get some context on why the above is the case, see the update to the answer at this link: https://math.stackexchange.com/a/3515756/708680
I would like to slightly reformulate the above expression so that instead of the aforementioned distances between points of tangency being powers of $\phi$ starting at $\phi^{-1}$ (for positive numbers) and increasing by powers of $\phi$ on alternate sides, the distance between 'wave-center' points are powers of $\phi$ starting at $\phi^{-1}$ (for positive numbers) and increasing by powers of $\phi$ on alternate sides, instead. Here 'wave-center' points are defined as points on the wave whose $y$ is half of the distance between any point where the curve crosses $y$ and the closest next point of crossing to that point.
See here for a graph showing the expression, points of tangency, 'wave-center' points, etc..., ['Wave-center' points are in red] (Please note that the 'wave-center' points in my graph are not the ones I want for the new expression but are instead just to show what I mean by 'wave-center' points): https://www.desmos.com/calculator/v7pmwr5oj9
I want the adjusted parametric equations to retain the following while being altered in the aforementioned manner; they should:


*

*0. Be of the form: $(x,y)=\big(f(t)^{-1}\cdot\sin(t), f(t)\big)$. For the sake of clarity I add that, for the original equations, this $f(t)$ was in the form 
$$
f(t)=\left(\alpha\cdot\phi^{t-{\pi/2}/\pi}+\beta-\frac{\alpha}{\phi^{1/2}}\right).
$$

*1. Start at $(0, 1)$ for positive and negative numbers.

*2. Have points of tangency to $x\cdot y=±1$ (as a result of 0.).

*3. Maintain a smooth, sinusoidal, 2-D spiral nature.

*4. Be written in terms of $\sin(t)$.


Thanks for your help.
 A: Adapting the argument from my previous answer to a related question (this time without changing orientation or shifting phases), we know that a curve parameterized by
$$(x,y) = \left(\frac{\sin t}{f(t)}, f(t)\right)\tag{1}$$
meets, and is tangent to, the hyperbolas $xy=\pm 1$ when $t$ is an odd multiple of $\pi/2$. It crosses the $y$-axis when $t$ is an even multiple of $\pi/2$; that is, when an integer multiple of $\pi$. Define $P_k = (x_k, y_k)$ where $t = k\pi$. We'll assume specifically that $t=0$ corresponds to the point $(0,1)$; for more generality, we'll take this to be $(0,\beta)$, so that we have
$$f(0) = \beta \tag{2}$$
OP defines a "wave center" as a point vertically halfway between two consecutive points where the curve crosses the $y$-axis. The $y$-coordinate of such a point is therefore $\frac12(y_k+y_{k+1})$ for some integer $k$. We seek the distances between alternate wave centers to be a power of $\phi$; again, for more generality (and to match OP's other related question), we'll take this to be a scaled power of $\phi$, giving this relation
$$\frac12(y_{k+2}+y_{k+3})-\frac12(y_{k}+y_{k+1})= \alpha \phi^{k-1} \tag{3}$$
where the power $k-1$ assures OP's desired value $\phi^{-1}$ for $k=0$. (Any index error can be reconciled by adjusting $\alpha$.)
Observing that
$$\phi^{k+3}+\phi^{k+2}-\phi^{k+1}-\phi^k = \phi^{k+3}+\phi^k\left(\phi^2-\phi-1\right) = \phi^{k+3} \tag{4}$$
(exploiting the golden ratio relation $\phi^2=\phi+1$), it's reasonable to suspect that our function has the form
$$f(t) = 2\alpha\phi^{t/\pi-4}+c \tag{5}$$
where $c$ is a constant that vanishes in $(3)$ but that we can recover from $(2)$:
$$\beta = f(0) = 2\alpha\phi^{-4}+c\tag{6}$$
Thus, we have

$$f(t) = \frac{2\alpha}{\phi^4}\left(\phi^{t/\pi} - 1 \right)+\beta  \tag{$\star$}$$

The curve parameterized by $(1)$ with $\alpha=\beta=1$ is as follows:

