2012 AIME II Problem 9

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

\frac{\sin 2x}{\sin 2y}

is just the product of the two given ratios

Square both given equations and add, using

\sin^2 + \cos^2 = 1,

to find

\cos^2 y;

then apply

\cos 2\theta = 2\cos^2\theta - 1

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.

Problem 9 in other years