Question

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 $p + q.$

Answer

Solution:

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

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 Problem 9

Solution:

Problem 9 in Other Years