For $\pi \le \theta < 2\pi$, let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$. Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Noticing the $\sin$ and $\cos$ in both $P$ and $Q,$ we think of the angle addition identities: 
\[\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b - \sin a \sin b\]
With this in mind, we multiply $P$ by $\sin \theta$ and $Q$ by $\cos \theta$ to try and use some angle addition identities. Indeed, we get 
\begin{align*} P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\ &= \cos \theta - \dfrac{1}{4} \cos \theta + \dfrac{1}{8} \sin{2 \theta} + \dfrac{1}{16} \cos{3 \theta} + \cdots \\ &= \cos \theta - \dfrac{1}{2}P \end{align*}
after adding term-by-term. Similar term-by-term adding yields \[P \cos \theta + Q \sin \theta = -2(Q - 1).\] 
This is a system of equations; rearrange and rewrite to get \[P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta\] and \[P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.\] Subtract the two and rearrange to get \[\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.\] Then, square both sides and use Pythagorean Identity to get a quadratic in $\sin \theta.$ Factor that quadratic and solve for $\sin \theta = -17/19, 1/3.$ Since we're given $\pi\leq\theta<2\pi,$ $\sin\theta$ is nonpositive. We therefore use the negative solution, and our desired answer is $17 + 19 = \boxed{036}.$