Let us consider a sphere of radius R and a ring element PQTS of thickness PQ on the surface of the shere.

OA is normal through the centre of the ring element.

Here $\angle{QOA}=\theta$ and the angular width of the ring element is $\angle{POQ}=d\theta$.

Let $r$ be the radius of the ring QT, so $r=R\ \sin\theta$.

Now the width of the ring element is $PQ=R\ d\theta$.

Therefore the area of the surface of the ring element is given by,

$dA=2\pi{r}\times{PQ}=2\pi{r}\times{R\ d\theta}$

$or,\ dA=2\pi{R\ \sin\theta}\times{R\ d\theta}$

$or,\ dA=2\pi{R^2}\ \sin\theta\ d\theta$

Now the solid angle subtended by the ring at the centre O of the sphere is given by,

$\displaystyle{d\omega=\frac{dA}{R^2}=\frac{2\pi{R^2}\ \sin\theta\ d\theta}{R^2}}$

$\displaystyle{or,\ d\omega=2\pi\ \sin\theta\ d\theta}$

Again, $\displaystyle{\sin\theta=\sin\ 2\frac{\theta}{2}=2\ \sin\frac{\theta}{2}\ \cos\frac{\theta}{2}}$

So the solid angle subtended by the circular ring element can be expressed as,

$\displaystyle{d\omega=4\pi\ \sin\frac{\theta}{2}\ \cos\frac{\theta}{2}\ d\theta}$