Editorial: fix algorithm of Math.atan2

[Josh-Cena]

Fix #2897.

I have no idea if this meets the conventions, but I hope the idea is at least correct. Below is the output of Math.atan2 in Node:

[bakkot]

Thanks for the PR, but isn't this wrong? Consider the case in your issue: y=1, x=-1. In that case we'll have r = atan(1/-1) i.e. r = -0.7853981633974483. Then it will fall into the nx < 0, ny > 0 case, which will do r = π - r, i.e. 3.9269908169872414. That's wrong; it should be 2.356194490192345.

I think this ends up simplest if we do the inverse tan of the absolute value of y/x, and then specify the quadrant explicitly. Basically

function atan2(y, x) {
  let r = Math.atan(Math.abs(y/x));
  if (x > 0 && y > 0) return r;
  if (x > 0 && y < 0) return -r;
  if (x < 0 && y > 0) return Math.PI - r;
  if (x < 0 && y < 0) return -Math.PI + r;
}

@Josh-Cena want to update the PR to work like that? Otherwise I'll get to it in the next week or two.

[Josh-Cena]

OK, I'll take a look. If I don't get to it in the next week or two then I probably will not ever :P

[Josh-Cena]

@bakkot I realized I only need to take abs for x (by referencing the diagram above). This way I avoid the extra negation when x > 0 && y < 0.

[bakkot]

@Josh-Cena That does work but I think it's significantly harder to understand than the version which just lists all four cases explicitly. Avoiding an extra negation in the spec algorithms isn't really a goal.