Before decay, the nucleus is in state with total angular momentum J and symmetry axis quantization M :

$\Phi_{JM}$

Say, the emitted radiation (can be EM wave or particle ) carries angular momentum  l and axis quantization m, its wavefunction is:

$\phi_{lm}$

then the daughter nucleus has angular momentum j and $m_j$, the wave function is

$\Psi_{j m_j}$

their relation is:

$\Phi_{JM} = \sum_{m, m_j}{\phi_{lm} \Psi_{j m_j} \left< l m j m_j | JM \right>}$

where the $Latex \left< l m j m_j | JM \right>$ is Clebsch-Gordan coefficient.

The wave function of the emitted radiation from a central interaction takes the form:

$\phi_{lm} = A_0 u_{nl}(r) Y_l^m(\theta,\phi)$

The angular distribution is:

$\int |\phi_{lm}|^2 dx^3 =A_0^2 \int |u_{nl}|^2 r^2 dr |Y_l^m|^2$

for a fixed distance detector, the radial part is a constant. Moreover, not every spherical harmonic contribute the same weight, there is weighting factor due to Clebsch-Gordan coefficient.   Thus, the angular distribution is proportional to:

$W(\theta) \propto \sum_{m_j=M-m}{|Y_l^m|^2 |\left|^2 }$

For example, JM=00, The possible (l, j) are (0,0), (1,1), (2,2) and so on, the $m=-m_j$. The C-G coefficient are,

$\langle 0000 | 00 \rangle = 1$

$\langle lml-m | 00 \rangle = \frac{1}{\sqrt{2l+1}}$

thus,

$Y_0^0 = \frac{1}{4\pi}$

$\displaystyle \sum_{m}{\left|Y_l^m\right|^2 \left|\langle l m l -m |0 0 \rangle \right|^2 } = \sum_m|Y_l^m|^2 \frac{1}{2l+1}= constant$

Thus, the angular distribution is isotropic.