Associativity and piecewiseness

Question

I have this confession: Piecewise definitions have always considerably annoyed me (with certain exceptions).

For $x,y\in (-1,1),$ let $x\circ y := xy -\sqrt{(1-x^2)(1-y^2)}\in(-1,1).$

(The part that says $\text{“}{\in}\text{”}$ is a simple exercise, not part of the definition.)

Proposision:

If $\arccos x + \arccos y + \arccos z < \pi\quad\longleftarrow \text{ piecewise}$
then $x\circ(y\circ z) = (x\circ y)\circ z.$

The way the proposition is stated probably makes it clear how to prove it: $$ x\circ y = xy-\sqrt{(1-x^2)(1-y^2)} = \cos\alpha\cos\beta-\sin\alpha\sin\beta = \cos(\alpha+\beta) $$ so it becomes mere associativity of addition. If the sum of the angles is more than a half-circle then we have a $\pm\sqrt{\cdots\cdots}$ issue, i.e. more piecewise stuff, and moreover we no longer have associativity.

So what is an elegant (in particular, short and without too much piecewise stuff) but completely elementary (say high-school-algebra-level) way to restate and prove the proposition without mentioning anything about trigonometric functions or otherwise constructing an isomorphism between this operation and another that everybody knows is associative?

(Maybe I would have answered this myself already if my distaste for piecewise definitions weren't making me gag. Maybe I'll post an answer below.)

Answer 1

Consider $S:=\{\,z\in\Bbb C\mid |z|=1\,\} $, show that $S$ is closed under multiplication, which is of course associative. Then for $x,y\in[-1,1]$ define $$x\circ y:=\Re (zw)\qquad \text{where }z,w\text{ are the unique elements of }S\text{ with }x=\Re z, y=\Re w, \Im z\ge 0,\Im w\ge 0$$ Quickly check that this makes $x\circ y={}$(your formula) and conclude that $(x\circ y)\circ z=x\circ (y\circ z)$ holds at least when the complex products in the definitions of $x\circ y$ and of $y\circ z$ have non-negative imaginary part.