For \pi \leq \theta<2 \pi, let
\begin{aligned}
P= & \frac{1}{2} \cos \theta-\frac{1}{4} \sin 2 \theta-\frac{1}{8} \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+\ldots
\end{aligned}
and
\begin{aligned}
Q= & 1-\frac{1}{2} \sin \theta-\frac{1}{4} \cos 2 \theta+\frac{1}{8} \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+\ldots
\end{aligned}
so that \frac{P}{Q}=\frac{2 \sqrt{2}}{7}. Then \sin \theta=-\frac{m}{n} where m and n are relatively prime positive integers. Find m+n.