Deriving Parametric Equations For A Hyperbolic PHI Sine-Wave This Question was formulated somewhat improperly: See here for the correct question:
Parametric Equations for A Logarithmic Sine-wave With Alternately Offset Points of Hyperbolic Tangency !!
Note. Please Read Everything Carefully and study the Graph Before Answering!
(PLEASE, See second fig. [the close up one] also for more detailed info about the geometry; this is how I think the most self-evident solution will graph but don't feel too constrained by it.)
I found the geometry for a sine wave that I've been trying to derive (and have posted about here before). (See my fig.) Everything appears pretty straightforward, however, I seem to lack the skill to derive it. My image should give you most of the information you need. But, here are a few points worth describing explicitly:

*

*The wave-length one side is always Phi to the one on the other side. (I know I'm not using the word 'wave-length' in a standard manner!) That is, if it's 1 on the right, it will be 1.618.... on the left (see fig.)


*I'm looking for parametric equations; they should take the form:
$x(t)=(FUNCTION)^{-1}*\sin(t), y(t)=(FUNCTION)$
I've given you $y(t)$ and $x(t)$ in the geometry of my image, because the whole thing should fit perfectly on $1/x$, thus the function for $x(t)$ is the one for $y(t)$ raised to the minus 1 or vice versa (Obviously the $\sin(t)$ itself is NOT raised to a negative power or such like!)


*If you need points for scale, I think the graph first crosses $y$ at $(0,1)$, in other words, it starts there in the same way that this: https://www.desmos.com/calculator/f53khj12ne Starts at $(0,1)$. I think it next crosses $y$ at $(0, 1.618\dots)$. These are guesses, so don't adhere to them if they don't make sense to you!


*Please try to find something where the function for the geometry is NOT inside the sine function ('$\sin(t)$'): I want to play with the curve and trying to get the expression out of the sine function might be a pain for me.
I can't wait to see this curve graphed! Thank you all so much; I'm very grateful for your cleverness and effort!

*
 A: I found something that seems to be heading in the right direction. However, does not really fit your curve nd doesn't uise PHI, but it might inspire you to think up the answer. i hope this helps: 
https://www.desmos.com/calculator/7bhjuv91c1
A: I'm going to re-orient things, and shift a phase, for reasons that I hope become clear.

A curve parameterized by
$$(x,y)=\left(f(t),\frac{\cos t}{f(t)}\right) \tag{1}$$
meets, and is tangent to, the hyperbola(s) $xy=\pm 1$ when $t$ is an integer multiple of $\pi$. Let $P_k = (x_k,y_k)$ be the point of tangency corresponding to $t = k\pi$.
(Note that the $P_k$ are not the local maxima and minima of the graph, since the tangent lines at those points are not horizontal.)
We want the horizontal offsets between every-other point of tangency to be a power of $\phi$; specifically, we want 
$$x_{k+1}-x_{k-1} = \phi^k \tag{2}$$
I suspect that OP intends the graph to bounce between the branches of the hyperbolas without crossing the $y$-axis (OP's $x$-axis). Moreover, it seems appropriate —but apparently, it is not; see "Update" below— for the graph to approach the $y$-axis, so that the $x$-coordinate of $P_0$ is the accumulated horizontal offsets in the sum
$$x_0 = \phi^{-1}+\phi^{-3}+\phi^{-5} + \cdots = \frac{\phi^{-1}}{1-\phi^{-2}}=\frac{\phi}{\phi^2-1}=\frac{\phi}{(\phi+1)-1} = 1 = \phi^0 \tag{3}$$
(where we have exploited the golden ratio property $\phi^2 - \phi - 1 = 0$). Likewise,
$$x_{-1} = \phi^{-2}+\phi^{-4}+\phi^{-6}+\cdots = \frac{\phi^{-2}}{1-\phi^{-2}}=\phi^{-1}\qquad\text{and}\qquad x_1 = 1 + x_{-1} = \phi^1 \tag{4}$$
Interesting. We have three instances where the subscript on $x$ matches the power on $\phi$. Well, if $x_{k-1}=\phi^{k-1}$, relation $(2)$ allows us to write 
$$x_{k+1} =x_{k-1}+\phi^k = \phi^{k-1}+\phi^{k} = \phi^{k-1}(1+\phi) = \phi^{k-1}\phi^2=\phi^{k+1} \tag{5}$$
so that, by induction, all subscripts on $x$ match the powers on $\phi$. We can extend this notion from integer $k$ to all reals by taking

$$f(k\pi) =x_k= \phi^k \quad\to\quad f(t) = \phi^{t/\pi}\quad\to\quad (x,y) = \left(\phi^{t/\pi},\phi^{-t/\pi}\cos t\right) \tag{$\star$}$$

This certainly seems to give the desired plot:


Update.
In comments below and in a revised question, OP updated the requirements so that (in my re-oriented context) the curve must pass through $(1,0)$; for greater generality, we'll make this $(\beta,0)$. Moreover, the revised question asks that the offsets between tangent points be scaled powers of $\phi$. These changes are not difficult to accommodate. Let's return to the above analysis at $(2)$, adjusting it to include $\alpha$:
$$x_{k+1}-x_{k-1} = \alpha\phi^k \tag{2'}$$
Observing that 
$$\phi^{k+1}-\phi^{k-1} = \phi^k \left( \phi - \frac{1}{\phi}\right) = \phi^k (\phi-(\phi-1)) = \phi^k \tag{3'}$$
it's reasonable to suspect that $f$ has the form
$$f(t) = \alpha\phi^{t/\pi}+c \tag{4'}$$
for some constant $c$ that gets lost in the difference in $(2')$. 
Previously, getting the curve to approach the $y$-axis amounted to having $c=0$ (with $\alpha=1$). Now, to pass through $(\beta,0)$, all we need to do is force $f(t)$ to be $\beta$ when $\cos(t)$ is $0$; specifically, OP wants the curve to meet $(\beta,0)$ between my $P_1$ and $P_{-1}$, so we take $t=-\pi/2$. Solving gives
$$\beta = f\left(-\frac{\pi}{2}\right) = \alpha\phi^{-\pi/2/\pi}+c \qquad\to\qquad c = \beta-\frac{\alpha}{\sqrt{\phi}} \tag{5'}$$
whence 

$$f(t) = \alpha\phi^{t/\pi} - \frac{\alpha}{\sqrt{\phi}} + \beta  \tag{$\star$'}$$

For $\alpha=\beta=1$, the plot is as follows:

The substitution $t\to t-\pi/2$ shifts the phase of things so that $(\beta,0)$ occurs at $t=0$. Moreover, it trades $\cos t$ for $\sin t$ in the parameterization, so that, calling the shifted function $f_0$, we have

$$f_0(t) = \alpha\phi^{(t-\pi/2)/\pi} + \beta - \frac{\alpha}{\sqrt{\phi}} = \frac{\alpha}{\sqrt{\phi}}\left(\phi^{t/\pi}-1\right) + \beta \quad\to\quad (x,y) = \left(f_0(t),\frac{\sin t}{f_0(t)}\right)$$

