I've been looking for a (formal) derivation of the following equation $\frac{\cos(\theta)dA}{r^2} = d\omega$. Where $d\omega = \sin(\theta_x)d\theta d\phi$ is the differential solid angle, and $dA$ is some oriented surface differential area element. $r$ is the distance to this element, and $\theta$ is the angle between the radius vector and the normal of the area element. There was already a similar question on here: Proof of $\cos(\theta) da=r^2 d\Omega$
Please refer to the image provided there for a clarification. However the thread linked above was marked as 'answered' even though no formal proof was provided. I have the intuition why this works, I just want to see the formal way one would prove this. For one thing all computer graphics books that use this fact always ignore the derivation, so I have been very interested in how one can prove it.
Edit: Found a formal proof: http://www.stewartcalculus.com/data/ESSENTIAL%20CALCULUS%20Early%20Transcendentals/upfiles/challenge/ess_cp_13_stu.pdf
We can start by thinking a differential surface ($dA$), which is oriented with an angle $\beta$ with respect to the normal of the surface of the sphere and this surface has area vector $d\vec{A}=dA\hat {n}_A$
And the normal of the sphere is $\hat {r}$
Hence, we can take the projection of $dA$ on to the sphere by $d\vec {A} \cdot \hat {r}$ since length of $\hat{r}$ is $1$ we can write, $$ d\vec{A} \cdot \hat {r} =dAcos(\beta)$$ So this is the area element that falls on the surface of the sphere. Lets call this area element $dA'$
Then from here we can write $$dA'/r^2=dw$$
The general derivation of $dA'/r^2=dw$ is fairly simple. The area element on the sphere can be calculated from the cross products of other two elements, so the area element $ds_r$ can be written as, $$d\vec{s}_r=d\vec{s}_{\theta} \times d\vec{s}_{\phi}$$ where $0<\theta<\pi$ and $0<\phi<2\pi$.
Here $d\vec{s}_{\theta} =rd\theta\hat {\theta}$ and $d\vec{s}_{\phi}=rsin(\theta)d\phi\vec {\phi}$ so we have,
$$d\vec{s}_r=rsin(\theta)d\phi\vec {\phi} \times rd\theta\hat {\theta}$$ $$d\vec{A'}=d\vec{s}_r=r^2sin(\theta) d\theta d\phi \hat {r}$$
or in magnitude, $$d{A'}=r^2sin(\theta) d\theta d\phi$$ and lets call $dw=sin(\theta) d\theta d\phi$ so we have $$dA'=r^2dw$$
but $dA'=dAcos(\beta)$ so we have
$dAcos(\beta)/r^2=dw$