DISCLAIMER - what follows is the simplified "analysis" that leads to the expression you were asking about. This is NOT in fact the correct way to treat this problem, as was pointed out in the comments.
When a blade with area $A$ moves through the air at a certain velocity $v$ and at an angle $\theta$, it "cuts through" an volume of air given by $V = A\sin\theta v$ every second - the projection of the area of the blade multiplied by the distance it moves in unit time.
This body of air is deflected downwards, and the velocity it attains is $v_d = v\sin\theta$. Now we can compute the mass of the body of air from the density and volume: $m = \rho V = \rho A \sin\theta v$. To accelerate a certain mass per unit time downwards, we need a force $F\Delta t = m\Delta v$. It follows that
$$\begin{align}F &= m\Delta v \\
&= \left(\rho A \sin\theta v\right)\left( v\sin\theta\right)\\
&=\rho A v^2 \sin^2\theta\end{align}$$
Now all you have to do is convert the linear velocity to a rotating velocity, noting that not every point on the blade will travel at the same speed when it's going around in a circle.
Note - lots of simplifying assumptions went into the above. The real physics of a blade moving through air has to take account of the fact that it's not only the air right in front of the blade that is moved... but the general expression (showing the linear relationship with density and area, the quadratic relationship with velocity, and the dependence on the square of the attack angle) looks just like the one you quoted. Of course when $\theta$ (or $\phi$, in the expression you quoted) gets too large, the air flow will "stall" and the force will get smaller, not larger. That shows the limitation of this simplistic approach (which only works over a limited range of angles and velocities, and with all the other simplifications already mentioned).
