Question

设 $x$ 和 $y$ 是实数，满足 $\frac{\sin x}{\sin y} = 3$ 且 $\frac{\cos x}{\cos y} = \frac{1}{2}$ 。数值 $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ 可表示为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质的正整数。求 $p + q$ 。

Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac{1}{2}.$ The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：107

知识点：三角恒等式代数变形

难度评级：2560

解答：

由二倍角公式， $\begin{aligned} \frac{\sin 2x}{\sin 2y} &= \frac{2\sin x \cos x}{2\sin y \cos y} \\ &= 3 \cdot \frac{1}{2} = \frac{3}{2}. \end{aligned}$

将已知方程平方，得 $\sin^2 x = 9\sin^2 y$ 和 $\cos^2 x = \frac{1}{4}\cos^2 y$ 。相加得到 $1 = 9(1 - \cos^2 y) + \frac{1}{4}\cos^2 y$ ，所以 $\frac{35}{4}\cos^2 y = 8$ ，并且 $\cos^2 y = \frac{32}{35}$ 。于是 $\cos 2y = 2\cos^2 y - 1 = \frac{29}{35}$ ，且 $\cos 2x = 2\cos^2 x - 1$ $= \frac{1}{2}\cos^2 y - 1$ $= -\frac{19}{35}$ ，所以 $\frac{\cos 2x}{\cos 2y} = -\frac{19}{29}$ 。

所求值为 $\frac{3}{2} - \frac{19}{29} = \frac{87 - 38}{58} = \frac{49}{58}$ ，所以 $p + q = 49 + 58 = 107$ 。

From the double-angle formula, $\begin{aligned} \frac{\sin 2x}{\sin 2y} &= \frac{2\sin x \cos x}{2\sin y \cos y} \\ &= 3 \cdot \frac{1}{2} = \frac{3}{2}. \end{aligned}$

Squaring the given equations, $\sin^2 x = 9\sin^2 y$ and $\cos^2 x = \frac{1}{4}\cos^2 y.$ Adding, $1 = 9(1 - \cos^2 y) + \frac{1}{4}\cos^2 y,$ so $\frac{35}{4}\cos^2 y = 8$ and $\cos^2 y = \frac{32}{35}.$ Then $\cos 2y = 2\cos^2 y - 1 = \frac{29}{35}$ and $\cos 2x = 2\cos^2 x - 1$ $= \frac{1}{2}\cos^2 y - 1$ $= -\frac{19}{35},$ so $\frac{\cos 2x}{\cos 2y} = -\frac{19}{29}.$

The requested value is $\frac{3}{2} - \frac{19}{29} = \frac{87 - 38}{58} = \frac{49}{58},$ and $p + q = 49 + 58 = 107.$

2012 AIME II 第 9 题

答案：107

难度评级：2560

解答：

其他年份的第 9 题