### Yay

2sin^{−1}(sqrt(2+(1−*P*)cos(*φ*_{1}+*φ*_{2})−(1+*P*)cos(*φ*_{1}−*φ*_{2}))/2)

where *P*=cos(*θ*_{1}−*θ*_{2}).

Phew.

**Edit:**

cos^{−1}(cos*φ*_{D}cos^{2}*θ*_{A}−cos*φ*_{S}sin^{2}*θ*_{A})

where *φ*_{S}=*φ*_{1}+*φ*_{2}, *φ*_{D}=*φ*_{1}−*φ*_{2} and *θ*_{A}=(*θ*_{2}−*θ*_{1})/2.

zqfmbgastralblueJust a bunch of trigonometric identities applied... XD

The second formula comes from the (forgotten) fact that the inner product of two vectors sharing the same origin equals to the cosine of the angle between them. Much easier/simpler than the brute-force approach I took to derive the first formula (which involves getting the distance between the two points). :D

niq