The sidereal day is a mean quantity, derived from many years of observations (it goes back to Halley times). But there are additional variations due to changes in the orbital parameters of the Earth. This changes are partly well known and predictable by celestial mechanics, but a significant part is unpredictable.
Google for a curve called Polar Motion. It depicts the movement of the north pole over the surface of the Earth through the years. Look for both the theoretical and the real one (build from measures). You will see the theoretical one is spiral-alike, but the measured "real" curve has a strongly erratic component superposed. It is even affected by earthquakes, and so is the exact position of the vernal equinox...
It has no sense to fine tune the determination of each sidereal day. It would be unrealistic, since the erratic, unpredictable contributions are in most cases bigger than the calculated corrections to the mean sidereal day.
Apart from that, I think there are some points in your reasoning that may be improved:
"Transit" is a well defined thing in astronomy. A line cannot transit. Moreover, the Sidereal Day is defined as the mean time between two transits of the Vernal Equinox. That point is, by definition, the zero of Right Ascension, always.
What you refer as "equatorial longitude" and "equatorial latitude" is very probably "Right Ascension" and "Declination", but I suggest that you carefully check that definitions, specially the sense (clockwise/counterclockwise) of your equatorial longitude vs Right Ascension, or you might probably run into a lot of problems when comparing your results with the literature.
In your calculations a factor cosĪµ appears and, although you may be doing something strictly correct from the geometrical point of view, I sincerely don't understand how you came to include the obliquity of the ecliptic in that calculations. You have probably projected some non constant movement from the ecliptic into the equator or something similar. Nobody can take you from doing that, but I think it leads unnecessarily to difficulties.
The ecliptic has nothing to do here. As long as we are not examining the movement of the real Sun, the ecliptic plays no role at all. It is the celestial equator the one that matters, since it is defined as the normal plane to the Earth rotation axis. There, in the celestial equator, all Hour Angles (and not "equatorial longitudes" or Right Ascensions, which are fixed) move with constant angular speed (as long as you neglect the 26000 yr periodic equinox precession and the regular + irregular movements of the north pole registered in the Polar Curve).
I kindly suggest you carefully learn the basic definitions and do lots of drawings. Thereafter, deriving relations will be safer.
Shouldn't it be relative to the Earth instead of the relative to the stars?
We need some reference background to plot the "movement" of the Sun. If we could see the stars during the day, and we were to go to a fixed point on the equator and mark the location of the Sun each day at noon on a star chart, this point would move in a circle through the stars once per year. The Sun rotates around the Earth more slowly than the stars do, so the number of solar rotations is one fewer than the number of sidereal rotations.
Imagine walking counterclockwise around a circular track, facing North the whole time. Suppose there's a light in the middle of the track. If you start out in the Eastern part of the track, the light will start out on your left. Once you get to the Northern part of the track, the light will be at your back. When you get to the Western part, it will be on your right. At the Southern part, it will be in front of you. So the light will appear to rotate around you counterclockwise.
So if the Earth didn't rotate at all, the Sun would appear to rise and set once over the course of the year. This one circuit due to the revolution around the Sun cancels out one of the 366 circuits due to the rotation of the Earth, leaving only 365 solar cycles.
Best Answer
According to the Wikipedia article you reference, "stellar day" is supposedly a new name for a planet's sidereal rotation period. However, I cannot find any documentation of this new name anywhere, and that includes my copy of Explanatory Supplement to the Astronomical Almanac, 3rd edition, edited by Urban and Seidelmann (University Science Books, 2012) which was just published within the month. This source is definitive and the term doesn't appear therein (okay at least not in the index). However, further digging found reference to it here
http://hpiers.obspm.fr/eop-pc/models/constants.html
but I've yet to find an actual statement of the change in terminology from "sidereal rotation period" to "stellar day" anywhere in the IERS conventions. Anywhere, the distinction is that the sidereal day is measured relative to the moving vernal equinox, which accounts for precession, whereas the sidereal rotation period (stellar day) is relative to the fixed inertial frame of background stars.