= Solution
The <Krivine rounding constant> has $t=c_K\pi/2=\log(1+\sqrt2)<\pi/2$. Since $|X_{ij}|\leq1$, the principal real <inverse sine> satisfies
$$
\arcsin(\sin(tX_{ij}))=tX_{ij}
$$
on every cross-block entry. The <matrix> $A$ has zero diagonal blocks, so those cross blocks are the only contributors to its <Frobenius inner product> with $\arcsin[Y]$. Explicitly,
$$
\operatorname{tr}(A\arcsin[Y])
=\sum_{i=1}^n\sum_{j=1}^mS_{ij}\arcsin(g(X_{i,n+j}))
=t\sum_{i,j}S_{ij}X_{i,n+j}
=t\operatorname{tr}(AX).
$$
No analogous identity is needed for the diagonal blocks involving $\sinh$.
Combining with the preceding <expectation> formula and optimality of $X$ gives the <bipartite sign rounding bound>
$$
\boxed{c_Kp_{\rm SDP}^*\leq v^*\leq p_{\rm SDP}^*},\qquad
\boxed{c_K=\frac2\pi\log(1+\sqrt2)=0.56109985\ldots}.
$$
At least one feasible rounded outcome attains at least this <expectation>. This is an expectation-based approximation guarantee for the specific bipartite sign problem; no claim is made that $c_K$ is the largest possible constant. The inequalities remain valid when the optimal objective is zero.
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