I've just started reading about the physics and features of Cosmic Microwave Background. I came across this pdg review titled Cosmic Microwave Background by D. Scott and G. F. Smoot says that
At every point in the sky, one observes a blackbody spectrum, with temperature $\color{blue}{T(\theta)}$.
Certainly, the satellites do not point "thermometers" at various points in the sky. However, I have a crude idea (may be wrong) about how they do it. Presumably, the detectors (in the satellites those map the temperature of the CMB sky) basically measure the flux of blackbody photons of various frequencies at a given direction $\hat{\textbf{n}}$ or equivalently, $(\theta,\phi)$. Then, from the intensity versus wavelength plot, read off the temperature $T$ of the CMB at each point.
The temperature $T(\theta,\phi)$ at any point on the CMB sky should be a function of that coordinate of the sky $(\theta,\phi)$. This is why one expands the temperature in terms of spherial harmonics as: $$T(\theta,\phi)=\sum\limits_{l,m}a_{lm}Y_{lm}(\theta,\phi).\tag{1}$$
However, in the reference mentioned above claims that the temperature becomes a function of $\theta$ only. In particular, it gives the following expression $$T(\theta)=T_0(1-\beta^2)^{1/2}(1-\beta\cos\theta)^{-1}\tag{2}$$ where $\theta$ is probably the angle between any photon momentum and the direction of uniform boost. Where did the azimuthal angle $\phi$ disappear?
Edit in response to @probably_someone's answer Formula (2), if I understood it correctly, is a special case of (1) when the anisotropy in temperature of CMB is solely due to a uniform motion of the observer w.r.t the CMB.
Does it mean that I can use equation (1), and set $\phi=0$ (without any loss of generality) to understand the effect of anisotropy caused by a uniform boost and read off the contributions at different orders in $l$? Is this the way to reconcile (1) and (2)?