Note that \begin{align*} P - Qi &= -i + \frac{1}{2} (\cos \theta + i \sin \theta) + \frac{1}{4} (-\sin 2 \theta + i \cos 2 \theta) + \frac{1}{8} (-\cos 3 \theta - i \sin 3 \theta) + \dotsb \\ &= -i + \frac{1}{2} (\cos \theta + i \sin \theta) + \frac{i}{2^2} (\cos \theta + i \sin \theta)^2 + \frac{i^2}{2^3} (\cos \theta + i \sin \theta)^3 + \dotsb \\ \end{align*}Let $ z = \cos \theta + i \sin \theta $. Then the sum above is an infinite geometric sum: \begin{align*} -i + \frac{z}{2} + \frac{iz^2}{2^2} + \frac{i^2 \cdot z^3}{2^3} + \dotsb &= \frac{-i}{1 - iz/2} \\ &= \frac{-2i}{2 - iz} \\ &= \frac{-2i}{2 - i (\cos \theta + i \sin \theta)} \\ &= \frac{-2i}{2 + \sin \theta - i \cos \theta} \\ &= \frac{-2i (2 + \sin \theta + i \cos \theta)}{(2 + \sin \theta)^2 + \cos^2 \theta} \end{align*}Matching real and imaginary parts, we get \begin{align*} P &= \frac{2 \cos \theta}{(2 + \sin \theta)^2 + \cos^2 \theta} \\ Q &= \frac{4 + 2 \sin \theta}{(2 + \sin \theta)^2 + \cos^2 \theta} \end{align*}Then from the equation $ \frac{P}{Q} = \frac{2 \sqrt{2}}{7}, $ \[\frac{\cos \theta}{2 + \sin \theta} = \frac{2 \sqrt{2}}{7}.\]Then $ 7 \cos \theta = 2 \sqrt{2} (2 + \sin \theta) $. Squaring both sides, we get \[49 \cos^2 \theta = 8 (2 + \sin \theta)^2,\]or $ 49 (1 - \sin^2 \theta) = 8 (2 + \sin \theta)^2 $. This simplifies to \[57 \sin^2 \theta + 32 \sin \theta - 17 = 0,\]which factors as $ (3 \sin \theta - 1)(19 \sin \theta + 17) = 0 $. Since $ \pi \le \theta < 2 \pi, $ $ \sin \theta $ is negative, so $ \sin \theta = \boxed{-\frac{17}{19}} $.