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:
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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.)
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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!
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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!
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Best Answer
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
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
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