I have a unit circle, and two angles: $\alpha=\angle{JON}\in[0,\pi]$ and $\beta=\angle{IOM}\in[0,\frac{\pi}{2}]$. Using angles, we can get points $N$, $M$ as on the image. Then, dropping a perpendicular from point $N$ to line $OM$ we can get point $A$.
Now, let us have another angle $\gamma=\angle{IOB}$.
The problem is how to calculate an angle $x=\angle{MAB}$ as a function of $\gamma$.
EDIT:
The original problem is mapping real rime cycle to astronomic time. Let the circle around $IOJ$ be the system of time representation. Time goes anticlockwise. Point $I=0$ (right) is 06:00, $J=0.5\cdot\pi$ (top) is 12:00, $-0.5\cdot\pi$ (bottom) is 00:00, $\pi$ is 18:00.
The sunrise equation gives us a sun noon, sunrise and sunset time for a current location. Let $M$ point be the noon time. We can set also $N$ point as sunrise time. Now, we can build a coordinate system in which $0$ will be real sunrise, $0.5\cdot\pi$ - real astronomic noon, etc.
Values $\in [0,\pi]$ will be the daytime, $[-\pi,0]$, resp. the nighttime.
I like MvG's idea of solution, but it seems to be unrealistic on the graphic: cardinality of the set of positive and negative values are the same, but should be different for chosen parameters $\alpha, \beta$.
As an illustration, cycle obtained using formula $\tan x = \frac{\sin(\gamma-\beta)}{\cos(\gamma-\beta)-\sin(\alpha+\beta)}$. Red line is original time cycle, cyan one is transformed cycle:
EDIT2:
After a little playing with coeffitients I have got such graphic:
The top picture represents a long day (summer), the bottom is a winter. I think, this solution looks adequately. Thanks @MvG.