2012 AIME II Problems/Problem 9

Problem 9

Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution

Examine the first term in the expression we want to evaluate, $\frac{\sin 2x}{\sin 2y}$ , separately from the second term, $\frac{\cos 2x}{\cos 2y}$ .

The First Term

Using the identity $\sin 2\theta = 2\sin\theta\cos\theta$ , we have:

$\frac{2\sin x \cos x}{2\sin y \cos y} = \frac{\sin x \cos x}{\sin y \cos y} = \frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}=3\cdot\frac{1}{2} = \frac{3}{2}$

The Second Term

Let the equation $\frac{\sin x}{\sin y} = 3$ be equation 1, and let the equation $\frac{\cos x}{\cos y} = \frac12$ be equation 2. Hungry for the widely-used identity $\sin^2\theta + \cos^2\theta = 1$ , we cross multiply equation 1 by $\sin y$ and multiply equation 2 by $\cos y$ .

Equation 1 then becomes:

$\sin x = 3\sin y$ .

Equation 2 then becomes:

$\cos x = \frac{1}{2} \cos y$

Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:

$1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$

Applying the identity $\cos^2 y = 1 - \sin^2 y$ (which is similar to $\sin^2\theta + \cos^2\theta = 1$ but a bit different), we can change $1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$ into:

$1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y$

Rearranging, we get $\frac{3}{4} = \frac{35}{4} \sin^2 y$ .

So, $\sin^2 y = \frac{3}{35}$ .

Squaring Equation 1 (leading to $\sin^2 x = 9\sin^2 y$ ), we can solve for $\sin^2 x$ :

$\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}$

Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$ , we can solve for $\frac{\cos 2x}{\cos 2y}$ .

$\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}$

$\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}$

Thus, $\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}$ .

Now Back to the Solution!

Finally, $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}$ .

So, the answer is $49+58=\boxed{107}$ .

2012 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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