Question 14
For $\pi \le \theta \lt 2\pi,$ let $$
\begin{aligned} P &= \frac{1}{2}\cos\theta - \frac{1}{4}\sin 2\theta \\ &\quad {}- \frac{1}{8}\cos 3\theta + \frac{1}{16}\sin 4\theta \\ &\quad {}+ \frac{1}{32}\cos 5\theta - \frac{1}{64}\sin 6\theta \\ &\quad {}- \frac{1}{128}\cos 7\theta + \ldots \end{aligned}
$$ and $$
\begin{aligned} Q &= 1 - \frac{1}{2}\sin\theta - \frac{1}{4}\cos 2\theta \\ &\quad {}+ \frac{1}{8}\sin 3\theta + \frac{1}{16}\cos 4\theta \\ &\quad {}- \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta \\ &\quad {}+ \frac{1}{128}\sin 7\theta + \ldots \end{aligned}
$$ 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 $m + n.$
\begin{aligned} P &= \frac{1}{2}\cos\theta - \frac{1}{4}\sin 2\theta \\ &\quad {}- \frac{1}{8}\cos 3\theta + \frac{1}{16}\sin 4\theta \\ &\quad {}+ \frac{1}{32}\cos 5\theta - \frac{1}{64}\sin 6\theta \\ &\quad {}- \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 \\ &\quad {}+ \frac{1}{8}\sin 3\theta + \frac{1}{16}\cos 4\theta \\ &\quad {}- \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta \\ &\quad {}+ \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 $m + n.$