Question

对 $\pi \le \theta \lt 2\pi$ ，令 $\begin{aligned} P &= \frac{1}{2}\cos\theta - \frac{1}{4}\sin 2\theta \\ &\quad {}- \frac{1}{8}\cos 3\theta + \frac{1}{16}\sin 4\theta \\ &\quad {}+ \frac{1}{32}\cos 5\theta - \frac{1}{64}\sin 6\theta \\ &\quad {}- \frac{1}{128}\cos 7\theta + \ldots \end{aligned}$ 且令 $\begin{aligned} Q &= 1 - \frac{1}{2}\sin\theta - \frac{1}{4}\cos 2\theta \\ &\quad {}+ \frac{1}{8}\sin 3\theta + \frac{1}{16}\cos 4\theta \\ &\quad {}- \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta \\ &\quad {}+ \frac{1}{128}\sin 7\theta + \ldots \end{aligned}$ 并满足 $\frac{P}{Q} = \frac{2\sqrt{2}}{7}$ 。若 $\sin\theta = -\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数，求 $m + n$ 。

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 $m + n.$

答案

答案：36

知识点：复数等比数列三角学二次方程

难度评级：3270

小提示：

把两个级数组合成 $Q + iP$ ：各项会成为一个公比为 $\frac{ie^{i\theta}}{2}$ 的等比级数。

Combine the two series as $Q + iP:$ the terms become a geometric series with ratio $\frac{ie^{i\theta}}{2}$

大提示：

求和可得 $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}$ 。令它等于 $\frac{2\sqrt{2}}{7}$ ，平方后保留满足 $\sin\theta \le 0$ 的根。

Summing gives $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}.$ Set it equal to $\frac{2\sqrt{2}}{7},$ square, and keep the root with $\sin\theta \le 0.$

解答：

符号以及正弦、余弦的交替提示我们使用 $i$ 的幂。实际上 $\begin{aligned} Q + iP &= 1 + \frac{1}{2}ie^{i\theta} + \frac{1}{4}i^2e^{2i\theta} \\ &\quad {}+ \frac{1}{8}i^3e^{3i\theta} + \cdots \\ &= \frac{1}{1 - \frac{ie^{i\theta}}{2}} = \frac{2}{2 - ie^{i\theta}} \end{aligned}\text{。}$ 因为 $2 - ie^{i\theta} = (2 + \sin\theta) - i\cos\theta$ ，乘以共轭数得到 $\begin{aligned} &Q + iP \\ &= \frac{2(2 + \sin\theta) + 2i\cos\theta}{5 + 4\sin\theta} \end{aligned}\text{，}$ 所以 $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}$ 。

令 $\frac{\cos\theta}{2 + \sin\theta} = \frac{2\sqrt{2}}{7}$ 并平方，得 $49(1 - \sin^2\theta) = 8(2 + \sin\theta)^2$ ，化简为 $\begin{aligned} &57\sin^2\theta + 32\sin\theta - 17 \\ &= (3\sin\theta - 1)(19\sin\theta + 17) \\ &= 0 \end{aligned}\text{。}$

因为 $\pi \le \theta \lt 2\pi$ 迫使 $\sin\theta \le 0$ ，所以 $\sin\theta = -\frac{17}{19}$ （此时 $\cos\theta = \frac{6\sqrt{2}}{19} \gt 0$ 与正的比值一致）。因此 $m + n = 17 + 19 = 36$ 。

The signs and the alternation between sines and cosines suggest powers of $i:$ indeed $\begin{aligned} Q + iP &= 1 + \frac{1}{2}ie^{i\theta} + \frac{1}{4}i^2e^{2i\theta} \\ &\quad {}+ \frac{1}{8}i^3e^{3i\theta} + \cdots \\ &= \frac{1}{1 - \frac{ie^{i\theta}}{2}} = \frac{2}{2 - ie^{i\theta}}. \end{aligned}$ Since $2 - ie^{i\theta} = (2 + \sin\theta) - i\cos\theta,$ multiplying by the conjugate gives $\begin{aligned} &Q + iP \\ &= \frac{2(2 + \sin\theta) + 2i\cos\theta}{5 + 4\sin\theta}, \end{aligned}$ so $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}.$

Setting $\frac{\cos\theta}{2 + \sin\theta} = \frac{2\sqrt{2}}{7}$ and squaring, $49(1 - \sin^2\theta) = 8(2 + \sin\theta)^2,$ which rearranges to $\begin{aligned} &57\sin^2\theta + 32\sin\theta - 17 \\ &= (3\sin\theta - 1)(19\sin\theta + 17) \\ &= 0. \end{aligned}$

Since $\pi \le \theta \lt 2\pi$ forces $\sin\theta \le 0,$ we get $\sin\theta = -\frac{17}{19}$ (and then $\cos\theta = \frac{6\sqrt{2}}{19} \gt 0,$ consistent with the positive ratio). Thus $m + n = 17 + 19 = 36.$

2013 AIME I 第 14 题

答案：36

难度评级：3270

小提示：

大提示：

解答：

其他年份的第 14 题