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 $p+q$.
The answer is $\boxed{107}$.
The first term in the target expression can be easily computed as $$\frac{\sin{2x}}{\cos{2y}}=\frac{2\sin{x}\cos{x}}{2\sin{y}\cos{y}}=3\cdot\frac{1}{2}=\frac{3}{2}$$--1
The second term can be transformed as $$\frac{\cos{2x}}{\cos{2y}}=\frac{2\cos^2x-1}{2\cos^2y-1}$$
We can solve $\cos^2x$ and $\cos^2y$ exactly by squaring the two given relationship: $$\frac{\sin^2x}{\sin^2y}=9\implies \frac{1-\cos^2x}{1-\cos^2y}=9\quad\text{and}\quad\frac{\cos^2x}{\cos^y}=\frac{1}{4}$$
This is a system of two equations and two variables. Its solution is $$(\cos^2x, \cos^2y)=\left(\frac{8}{35}, \frac{32}{35}\right)\implies\frac{\cos{2x}}{\cos{2y}}=-\frac{19}{29}$$
It follows that $$\frac{\sin{2x}}{\sin{2y}}+\frac{\cos{2x}}{\cos{2y}}=\frac{3}{2}-\frac{19}{29}=\frac{49}{58}$$