arccos_one_div_sqrt_two

In 4D Regge calculus on the integer lattice, each hinge carries a deficit angle built from dihedral angles of the incident 4-simplices. For the seed triangle hinge ${0,e_0,e_0+e_1}$, the flat (zero-curvature) Gram data produce two cosine values: $1/\sqrt{2}$ on the opposite-edge orbit and $0$ on the orthogonal orbit.

This module enumerates the full periodic Freudenthal star (six incident 4-simplices in four unit cubes) and converts those cosines into angles before summing. The present lemma is the elementary inverse-cosine evaluation needed for the opposite-edge seed angle.

The only upstream arithmetic facts used are the multiplicative unit laws $1\cdot x=x$ (in the foundation layers), applied inside the algebraic rewrite $1/\sqrt{2}=\sqrt{2}/2$.

Recall $\cos(\pi/4)=\sqrt{2}/2$. Show $1/\sqrt{2}=\sqrt{2}/2$ by clearing denominators: multiply both sides by $\sqrt{2}\cdot 2$, use $\sqrt{2}\neq 0$ and $2\neq 0$, then $\sqrt{2}\cdot\sqrt{2}=2$. Rewrite the goal, substitute the cosine identity, and apply $\arccos(\cos\theta)=\theta$ with $\theta=\pi/4\in[0,\pi]$ (positivity of $\pi/4$ and the bound $\pi/4\le\pi$).

Downstream, flatAngleSeedOpp_eq is the one-line identification of the opposite-edge flat seed angle with $\pi/4$, obtained by applying this lemma. That angle is one of the two values in the flat cosine multiset (four simplices at $1/\sqrt{2}$, two at $0$) that the module uses to prove the star angle sum equals exactly $2\pi$ and to assemble the full-star deficit class kernel on classes $(2,3,6,7,10,11,14)$.

Within the QG campaign this is a kernel-checked increment after the dihedral cosine calculus: it does not yet close Hessian assembly, Einstein–Hilbert convergence, or gap-action recovery, but it removes an elementary real-analysis obligation from those later steps.