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The spherical coordinate system has three coordinates:

the distance from the origin, denoted $\rho$;

the angle measured from the positive $xz$-plane, denoted $\theta$;

the angle from the positive $z$-axis, denoted $\phi$.

One can see that the idea is fairly similar to polar coordinates, since the distance to the origin is a coordinate. However, since we are now in 3-D space, we need 2 angle measures instead of just 1 to specify a point.

Notice that the angle $\theta$ is the same as in cylindrical coordinates. This makes conversion between spherical and cylindrical coordinates fairly easy to derive. Here are the conversion formulas: $$\rho^2=r^2+z^2$$ $$\tan\phi=\frac{r}{z}$$ $$\theta=\theta$$ $$r=\rho\sin\phi$$ $$z=\rho\cos\phi$$

Taking the conversion a step further, we can convert between spherical and Cartesian coordinates. These involve a little bit more computation, but they are still easy to prove. Here are the conversion formulas: $$\rho^2=x^2+y^2+z^2$$ $$\tan\theta=\frac{y}{x}$$ $$\tan^2\phi=\frac{x^2+y^2}{z^2}$$ $$x=\rho\sin\phi\cos\theta$$ $$y=\rho\sin\phi\sin\theta$$ $$z=\rho\cos\phi$$

Unfortunately, as in the polar coordinate system, there are things to watch out for. Every point has infinitely many representations. In two dimensions, it was bad enough. Think about this for a moment: $$(\rho,\theta,\phi)=(-\rho,-\theta,\phi)=(\rho,\theta+2\pi,\phi)=(\rho,\theta,\phi+2\pi)=(-\rho,\theta,\phi+\pi)=(\rho,-\theta,-\phi)$$ To get around this issue, we generally impose the following restrictions: $$\rho\ge 0$$ $$0\le\theta\lt2\pi$$ $$0\le\phi\le\pi$$ However, just as in polar coordinates, one must think beyond the typical range in some unusual circumstances.

There is also the issue of undefined coordinates. Points on the $z$-axis don't have a well-defined $\theta$, while the origin does not have a well-defined $\phi$.

Whenever one has a shape related to spheres or cones, one should definitely consider spherical coordinates. They are good for anything with spherical or even axial symmetry.

In astronomy, the spherical coordinates used involve Right Ascension (RA) and Declination (Dec). These are analogous to longitude and latitude respectively. Let us denote RA as $\alpha$ and Dec as $\delta$. Mathematically, $\alpha \in [0, 2\pi]$ and $\delta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ where those angles are in radians. One thing to note is that $\delta = -\frac{\pi}{2}$ corresponds to the south pole and $\delta = \frac{\pi}{2}$ corresponds to the north pole.

Okay, with that in mind, let's do a problem. Derive the formula for angular distance between the points $(\alpha, \delta)$ and $(\alpha_0, \delta_0)$.