Is there a source for this equation for a Bell's Theorem test? In this, he gives an example of entangled photons spin being measured – one photon sent to a detector at $0^\circ$, and the other photon sent to a detector at $20^\circ$. He says the odds of both detectors triggering a positive result is $5.8\%$, and he gives this as the formula for calculating that:
$$\frac12\sin^2 (20^\circ)$$
I would like to find an external source for this function, but there isn't one provided in that article. I think the example he gives in this article is really brilliant and illustrative and I would like to use a similar example, I just need a reliable source for this piece of it.
 A: The idea is that if the photon spin is "$0^\circ$ then it has a probability $\cos^2\theta$ of being transmitted through a detector oriented at "$\theta$". Photon spin is really known as polarization, with "$0^\circ$" being labeled as a horizontally polarized photon and $90^\circ$ being labeled as a vertically polarized photon. We can write the states as $|H\rangle$ and $|V\rangle$.
A general state (well, almost the most general but disregarding complex numbers) can be expanded in this basis as
$$|\psi\rangle=\cos x|H\rangle+\sin x|V\rangle$$ for some value of $x$. The likelihood of it passing through a detector oriented at $\theta$ is $\cos^2(x-\theta)$. A horizontally polarized photon has $x=0^\circ$, so the probability of transmission is $\cos^2\theta$ as desired, which goes to zero when the detector is perpendicular to this incident photon.
Now, for the entangled state in question (known as a singlet state), we have a state that is a superposition of two different combinations for the respective polarizations of each photon (I label the photons by $1$ and $2$):
$$|\mathrm{singlet}\rangle=\frac{|H\rangle_1 |V\rangle_2-|V\rangle_1|H\rangle_2}{\sqrt{2}}.$$ When we send the first photon of this joint state through a detector oriented at $0^\circ$, only the branch ${|H\rangle_1 |V\rangle_2}$ of the superposition can make it through. The photons have the same probability of being in this branch as in the other branch, so the probability thus far is $50\%$. Next, the second photon needs to make it through a detector (polarizer) oriented at $\theta$. The second photon is in state $|V\rangle$, which is like $|\psi\rangle$ with $x=90^\circ$. The probability of success for this second photon is thus $\cos^2(90^\circ-\theta)=\sin^2\theta$, which can be proven using trigonometry.
Putting these together, the joint probability is
$$P=\frac{1}{2}\times \sin^2\theta$$ as desired.
