Does accelerating cosmological expansion increase beam spread? In the standard textbook case, a transmitter of diameter $D$ can produce an electromagnetic beam of wavelength $\lambda$ that has spread angle $\theta=1.22\lambda / D$. But what happens in an expanding cosmology, especially one that accelerates so that there is an event horizon? Does $\theta$ increase with distance?
Obviously each photon will travel along a null geodesic and after conformal time $\tau$ have travelled $\chi=c\tau$ units of co-moving distance. The distance between the beam edges would in flat space be growing as $\delta=2c\sin(\theta/2)\tau$. Now, co-moving coordinates are nice and behave well with conformal time, so I would be mildly confident that this distance is true as measured in co-moving coordinates. 
But that means that in proper distance the beam diameter is multiplied by the scale factor, $a(t)\delta$ (where $t$ is the time corresponding to $\tau$), and hence $\theta$ increases. However, the distance to the origin in these coordinates has also increased to $a(t)(ct)$, so that seems to cancel the expansion - if we measure $\theta(t)$ globally by dividing the lengths. 
But it seems that locally we should see the edges getting separated at an accelerating pace; after all, the local observers will see the emitter accelerating away from them, producing a wider and wider beam near their location since it was emitted further away. From this perspective as time goes by the beam ends up closer and closer to $\theta=\pi$ (and ever more red-shifted, which presumably keeps the total power across it constant).
Does this analysis work, or did I slip on one or more coordinate systems?
 A: The first part of the answer is very simple, and you've worked it out yourself already. At a comoving distance $d_C$, the comoving diameter of the beam is simply $\delta_C = 1.22\lambda d_C/D$. If the beam is emitted at $t=t_0$, then at time $t$ the proper diameter of the beam is:
$$\delta = \frac{a(t)}{a(t_0)}\delta_C$$
Now the slightly tricky part. Talking about the spread angle at the location of the observer doesn't really make sense - any reasonable treatment of the source by the observer considers it a point source (unless the detector is larger than the beam; good luck with that!), and I don't think there's any local measurement that can determine the spread angle. The only reasonable question I can think to ask is "what is the apparent angular size of the beam as it would be measured from the point of emission"$^1$. This is a standard problem, and the answer is simply stated in terms of the angular diameter distance $d_A$ as $\theta_{\rm apparent}= \delta/d_A$. From here it's merely an exercise to pick a cosmology, a time of emission and e.g. a comoving distance then work out the relevant times, scale factors and so on.
Angular sizes in expanding metrics can do some funny things - for instance, in something like Planck cosmology (~our Universe) an object of fixed size subtends a smaller and smaller angle as it is placed further away, as your intuition expects... until it gets past a threshold distance, after which it starts to appear bigger again! This can be intuitively understood since a $1\,{\rm m}$ object at $t\sim0$ occupies a space corresponding to enormous present-day scales, and so it should nearly fill the sky when observed. In your case of an accelerating expansion, though, the end result is that the apparent angular size of the beam increases continuously.

$^1$ I guess you could also measure it easily from anywhere along the direction of travel, but from the point of emission seems to be the most interesting case.
