= Solution
Let $z=\cos\theta^\star$. Rotating the spin-one intermediate state between initial and final axes gives <probability amplitudes> proportional to $d^1_{\pm1,\pm1}$ for equal <fermion> <helicities> and $d^1_{\mp1,\pm1}$ for opposite ones. Thus, apart from a common positive <normalization>,
|| Channel | Angular weight \\
| $LL$ | $|g_L^eg_L^f|^2(1+z)^2$ \\
| $RR$ | $|g_R^eg_R^f|^2(1+z)^2$ \\
| $LR$ | $|g_L^eg_R^f|^2(1-z)^2$ \\
| $RL$ | $|g_R^eg_L^f|^2(1-z)^2$
The common factor $1/4$ from the <rotation> elements is absorbed into the <normalization>. Distinct external <helicities> are <Orthogonality>[orthogonal], so an unpolarized rate adds their squared <probability amplitudes> incoherently.
\b[The source interchanges the conventional left and right <electron> <Gauge coupling>[coupling] labels.] With $g_L=T_3-Q\sin^2\theta_W$ and $g_R=-Q\sin^2\theta_W$, the physical <electron> values are
$$
g_L=-\frac12+\sin^2\theta_W,\qquad g_R=\sin^2\theta_W.
$$
At $\sin^2\theta_W=0$, for example, the left <electron> still carries <weak isospin> $-1/2$, while the right <electron> is a <Singlet representation>[weak singlet]; the printed assignment gives the opposite result. This fixes how the helicity-channel table should be interpreted.<Electroweak Model and Constraints on New Physics by J. Erler and A. Freitas (2024)>[§10.1: <fermion> <Neutral current>[neutral-current] interaction and weak-isospin assignments]{ref}
For a <muon> final state, <lepton universality> gives the same pair of <Gauge coupling>[couplings]. Crucially, swapping both labels does not change the summed distribution. Put $L=g_L$, $R=g_R$; then
$$
\boxed{\frac{d\sigma}{dz}=\mathcal N\left[(L^4+R^4)(1+z)^2+2L^2R^2(1-z)^2\right]
=\mathcal N\left[(L^2+R^2)^2(1+z^2)+2(L^2-R^2)^2z\right].}
$$
Here $\mathcal N$ includes the resonance <propagator>, initial <spin> average and common <Gauge coupling>[coupling] <normalization>. Equivalently,
$$
\boxed{\frac1\sigma\frac{d\sigma}{dz}=\frac38(1+z^2)+\frac34\mathcal A_\ell^2z,\qquad
\mathcal A_\ell=\frac{L^2-R^2}{L^2+R^2}.}
$$
The source's simultaneous label interchange reverses $\mathcal A_\ell$ but leaves its square unchanged.
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