Trig Determinants

Clearly, $|A| + |B| = \cos(x+y) + \sin(x+y).$ If we let $u = \cos(x+y)$ , then we obtain:

$u + \sqrt{1-u^2} = \frac{1}{3}$ ;

or $1-u^2 = (1/3 - u)^2;$

or $1-u^2 = \frac{1}{9} - \frac{2}{3}u + u^2;$

or $0 = -8 - 6u + 18u^2;$

or $u = \frac{6 \pm \sqrt{6^2 - 4(18)(-8)}}{2(18)};$

or $u = \frac{1 \pm \sqrt{17}}{6} = \cos(x+y) \Rightarrow \sin(x+y) = \sqrt{1 - (\frac{1 \pm \sqrt{17}}{6})^2} = \frac{1 \mp \sqrt{17}}{6}$ ;.

Finally, $\sin(2x+2y) = \sin 2(x+y) = 2\sin(x+y)\cos(x+y) = 2(\frac{1 \mp \sqrt{17}}{6})(\frac{1 \pm \sqrt{17}}{6}) = \frac{1-17}{18} = -\frac{8}{9} = \frac{a}{b},$ and $a \cdot b = \boxed{-72}.$

If |A| + |B| = 1/3, this means that cos(x)cos(y) - sin(x)sin(y) + sin(x)cos(y) + cos(x)sin(y) = 1/3. Thus, cos(x+y) + sin(x+y) = 1/3.

Squaring both sides yields cos(x+y)^2 + sin(x+y)^2 + 2sin(x+y)cos(x+y) = 1/9. By the Pythagorean Identity, cos(x+y)^2 + sin(x+y)^2 = 1. Also, 2sin(x+y)cos(x+y) = sin(2 * (x+y)) = sin(2x+2y). Thus, sin(2x+2y) = -8/9, so -8 * 9 = -72 which is the final answer.