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Yeah, I know, "launch into target orbit". But let's assume we need to perform the maneuver.
A satellite in LEO equatorial orbit would need excessive amounts of delta-V to enter polar orbit. The amount needed in a very high orbit, on the other hand, is very low. Thing is getting from low to high orbit is costly, so the answer probably lies somewhere in the middle. So - what would be the optimal maneuver of turning LEO 90 degrees?
Let's take a look at this. The fuel usage, from Wikipedia, is
$\Delta{v_i}= {2\sin(\frac{\Delta{i}}{2})\sqrt{1-e^2}\cos(w+f)na \over {(1+e\cos(f))}}$
Assuming a 0 e orbit to begin with, this basically becomes $sqrt(2)*v$, where v is the orbital velocity. Thus, for a LEO velocity of around 7.8 km/s, that adds up to be quite a lot of fuel!
I believe the best strategy would be to do a nearly escape elliptical orbit, set so the $A_n$ or $D_n$ is at the peak of that elliptical orbit. When you reach the appoasis, your velocity will be very small, you can easily do the inclination switch there for minimal delta v. Then when you come back you give yourself the correct eccentricity. This will cost a significant amount of fuel still, but still reasonable. The escape velocity is about $sqrt(2)-1$ , so doubling that will give you an effective fuel budged of something less than your orbital velocity. The trick is to make the inclination change when your orbit is almost not moving, and that works best at extreme orbital heights on highly elliptical orbits.
Of course, I'm pretty confident you could do the same thing by a very clever Lunar flyby, much quicker. The delta v to do a lunar flyby from LEO is about 4.1 km/s, each way.
