The formula should be, at least for the first quadrant, $0 \leq \alpha < \pi/2$:
$$\beta=2 \ atan \left( \dfrac{e}{f}\right) \ \ (1)$$
where $e$ is the length of SR i.e.,
$$e=\sqrt{r^2-(r \sin \alpha - d)^2}-r \cos \alpha \ \ (2)$$
and
$$f=2 r \sin \alpha - d \ \ (3)$$
Why all that ? Let $P'$ be the symmetrical point $P'$ of $P$ with respect to x-axis, and $e=SR$ as said before.
the coordinates of $R$ being $(r \cos \alpha + e, r \sin \alpha -d)$, we have to express that $R$ belongs to the circle with radius $r$ by writing $(r \cos \alpha + e)^2+( r \sin \alpha -d)^2=r^2$, out of which we get (2).
the fact that $PP'=2 r\sin \alpha$ is clear, thus $P'S=2 r\sin \alpha -d$ whence (3).
the central angle theorem (http://www.mathopenref.com/arccentralangletheorem.html) is the last ingredient. Explanation: the angle from which arc PR is seen from point $P'$ is $\beta/2$, the half of angle $\beta$ from which the same arc PR is seen from the center of the circle. This angle $\beta/2$ is such that, in triangle $P'SR$, $\tan(\beta/2)=SR/SP'=e/f$, explaining formula (1).
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
$$f_0(t) = \alpha\phi^{(t-\pi/2)/\pi} + \beta - \frac{\alpha}{\sqrt{\phi}} = \frac{\alpha}{\sqrt{\phi}}\left(\phi^{t/\pi}-1\right) + \beta \quad\to\quad (x,y) = \left(f_0(t),\frac{\sin t}{f_0(t)}\right)$$
Best Answer
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
The curve parameterized by $(1)$ with $\alpha=\beta=1$ is as follows: