Ok, you are right, that was my mistake!

From now on, the angle is between r0 and the positive x-axis, in the same way that the distances r1 and r2 are calculated using the polar cooridnates of point P (Rcos(a0), Rsin(a0)).

I tried some more calculations, this time using my old formula with the defination of a0 as stated above:
a0 = 60 deg --> phase diff = 1/2pi rad

Now with your formula it would be:
a0 = 60 deg --> phase diff = pi*sin(90-60) = pi*sin(30) = 1/2pi rad

It's also the same for 45 deg
So it seems to be the same for 0, 45, 60 and 90 deg.

I'm pretty confident now that my long formula is exactly the same as pi*sin(a0)
Since, my formula also uses the radius R, but whatever radius I use I get the same results so it is actually independent of R, making me think there is some simplification I'm not seeing.



After using the fact that the phase difference = 2pi*f*t
and t = (r1-r2)/v
and f = v/wavelength,

I got this formula for the phase difference:

(formula 1)
R is the radius as shown in the last image.
The angle theta is the angle between the positive x-axis and r0.
and lambda = wavelength

So it seems now that:
formula 1 = pi*sin(90 - theta)


Can anybody show me that? I have tried to simplify my formula but it gets me nowhere...