This video is the best derivation I found: https://www.youtube.com/watch?v=a-dIl1Y1aTs.
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A single complex number $z = a + ib$ has two degrees of freedom, the two real numbers a and b.
A qubit, or a two-level quantum state can be represented with two complex numbers, the amplitudes of the $\ket{\psi} = \alpha\ket{0} + \beta\ket{1}$.
But $\alpha$ and $\beta$ are not unconstrained parameters.
First, we have the normalization constraint, which means that $\lvert \alpha \rvert^2 + \lvert \beta \rvert^2 = 1$.
So we may write $\ket{\psi}$ as:
$$\ket{\psi} = {\rm e}^{i\phi_0}\cos{\frac{\theta}{2}}\ket{0} + {\rm e}^{i\phi_1}\sin{\frac{\theta}{2}}\ket{1}$$
with $0 \le \theta \le \pi$ (since only positive real parts of the amplitude suffice), and $0 \le \phi_0, \phi_1 \lt 2\pi$ (since $2\pi$ is the same phase as $0$). Using $\frac{\theta}{2}$ is more convenient than $\theta$ for visualization.
Secondly, the global phase of a qubit does not matter – only the relative phase between the amplitudes is physically detectable. This means:
$$\ket{\psi} = {\rm e}^{i\phi_0}\cos{\frac{\theta}{2}}\ket{0} + {\rm e}^{i\phi_1}\sin{\frac{\theta}{2}}\ket{1} = {\rm e}^{i\phi_0}\left(\cos{\frac{\theta}{2}}\ket{0} + {\rm e}^{i\left(\phi_1-\phi_0\right)}\sin{\frac{\theta}{2}}\ket{1}\right)$$
Ridding the global phase factor of ${\rm e}^{i\phi_0}$, and writing $\phi = \phi_1 - \phi_0$, this state is equivalent to:
$$\cos{\frac{\theta}{2}}\ket{0} + {\rm e}^{i\phi}\sin{\frac{\theta}{2}}\ket{1}$$
So now we are able to describe a two-level quantum state with only two real numbers, $0 \le \theta \le \pi$ and $0 \le \phi \lt 2\pi$.
For $\ket{\psi} = \ket{0}$, $\theta = 0$, and for $\ket{\psi} = \ket{1}$, $\theta = \pi$ and $\phi = 0$.
While we could scatter plot qubit states in a 2-D $\theta {\rm vs.} \phi$ chart, visualizing using spherical co-ordinates gives a richer and more useful picture. In this spherical co-ordinate system since we have only unit vectors, we can render qubit states as points on the unit sphere with $\ket{0}$ on the north pole, and $\ket{1}$ on the south pole of the sphere.