FWL: Before this intrusion of ineptitude risks the moral endangerment of any further tender young minds, I feel obliged to point out the logical inequivalence between the statements cos(rational * 2pi) = algebraic ! and arccos(rational) = algebraic * 2pi ? the second of which is in general (to put it mildly) problematic. Those remaining in doubt about the matter should reflect upon the kindergarten fact arccos(1/3) = dihedral angle of regular tetrahedron, the expression of which as an algebraic angle ought to be worth a Fields at least! --WDS: I'd never claimed these 2 statements equivalent. Since FWL seems confused, let me clarify. Or repeat what I'd said before. Hopefully this also will unconfuse others such as RWG and "tender young minds." (Although, frankly, I think the problem lies more in cranky old minds.) FACT: Suppose you have an equation of the form FINITESUM_j A_j * arccos(B_j(u)) = C*pi where C and the A_j are integers, and the functions B_j(u) are algebraic functions of u. Then: u is algebraic. BECAUSE: You can rewrite the equation as realpart PRODUCT_j Y_j(u)^(A_j) = 0 where the Y_j(u) are complex numbers with unit norm that are algebraic functions of u, having realpart Y_j(u) = B_j(u). Here by "algebraic function" F(u) I mean "if u algebraic then F(u) is, and if y algebraic with F(u)=y, then u is." In gear applications, the fact the A_j are integers is related to the fact gear tooth counts are integers. Are we clear now? This same trick is also how one can prove (or disprove) any of my so-called "arccos miracles" in an automated fashion, and note, this proof NOWHERE INVOLVES trig or arctrig, which was the whole point of the trick -- the arctrig intentionally gets removed from the scene, allowing simpler pure-algebraic simplifiers to do their thing. Be happy.