 Solution

With $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$, the matrix of $S_z\cos \theta +S_x\sin \theta$ is $(\hslash /2)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right)$. Multiplication by $(c,s{)}^T$ and $(-s,c{)}^T$ gives eigenvalues $+\hslash /2$ and $-\hslash /2$, using the double-angle identities.

The one-particle transformation is the matrix $R=\left(\begin{array}{cc}c& -s\\ s& c\end{array}\right)$, with determinant one. The antisymmetric tensor $|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩$ transforms by this determinant under $R\otimes R$, so the singlet is unchanged. The total $S_z$ kills it. Its invariance under these rotations implies that total $S_z\cos \theta +S_x\sin \theta$ also kills it for every $\theta$, so total $S_x$ kills it. The spin commutator then gives total $S_y$ acting as zero. Therefore $S^2|\chi ⟩=0$ and it has total spin zero.

Rotating only particle B producesThe measurement probabilities are thereforeEach arrow represents the corresponding value $\pm \hslash /2$. The sequential measurements commute because they act on different particles. The two parallel outcomes leave triplet states with total spin one and $S^2=2{\hslash }^2$. Each antiparallel product state is a superposition of the spin-zero singlet and the spin-one zero-projection triplet, and is not an $S^2$ eigenstate. Outcomes of zero probability at a special angle do not actually occur.