Geocentric latitude and longitude of sub-satellite point

Question

I was trying to derive the expressions for the geocentric latitude and longitude of the sub-satellite point using Keplerian elements. The final equations connecting the combination of the Keplerian elements with the geocentric latitude and longitude doesn't contain the eccentricity and semi-major axis, and I was wondering why? I was also wondering what transformation equations to use to derive the expressions for the geocentric latitude and longitude. Any leads would be highly appreciated.
The starting variables that I used are the classical orbital elements, and I would like to arrive at:

1. geocentric latitude: $ \ \ \sin \phi = \sin i \sin(\omega + f)$
2. geocentric longitude: $ \ \ \tan(\lambda - \Omega) = \cos i \tan(\omega + f) $

Here $i$ is inclination, $\omega$ is the argument of perigee, $\Omega$ is the longitude of ascending node and $f$ is the true anomaly.

Answer 1

I can answer the question of why Semimajor Axis and Eccentricity aren't included in the calculation, at least.

What you have here are the main Keplerian Orbital Elements that define the plane of your orbit around Earth. Every orbit with the same inclination and Longitude of the ascending node lies in exactly the same plane, which intersects the globe of the Earth on a great circle.

Every orbit in that plane, with the same above elements. with the same Argument of Perigee has its Periapsis lined up along the same line drawn from the center of the Earth.

And True Anomaly is an angle measured in the plane of the orbit, from the Perigee to the current position of the spacecraft, along the direction of travel around the orbit.

As a result with those four elements fixed, seen from the center of the Earth, every object in orbit with the same Longitude of the Ascending Node, Inclination, Argument of Perigee, and True Anomaly at the same time is on the same line drawn from the center of the Earth, regardless of Semimajor Axis or Eccentricity.

This results in them all having the same surface Longitude and Latitude.

Answer 2

Source: PAGE 127 of Karttunen, Hannu., Pekka. Kröger, Heikki. Oja, Markku. Poutanen, Karl Johan. Donner, and SpringerLink. Fundamental Astronomy. Fifth ed. 2007. Web.

(We must be careful here; the equation for tan(λ−Ω) allows two solutions. If necessary, a figure can be drawn to decide which is the correct one.)

Implementation:

function [lat lon] = keplar2ll(incl,argp,RAAN,nu)
%% OF SUBSATELLITE POINT at epoch
% incl: inclination
% argp: argument of perigee
% RAAN: longitude of ascending node
% nu: true anomaly
lat = asind(sind(incl)*sind(argp+nu));
L = atand(cosd(incl)*tand(argp+nu));
if lat >=0
    if L>0
        lon = L + RAAN - 360;
    else
        lon = L + RAAN - 180;
    end
else
    if L>0
        lon = L + RAAN - 180;
    else
        lon = L + RAAN;
    end
end