Trigonometry Problems 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 $p + q$.
My student left me with a list of questions just now (he also said that he wanted to be helped at school :P)

\begin{align}
{\sin2x\over\sin2y}+{\cos2x\over\cos2y}=
{}& {2\sin x\cos x\over2\sin y\cos y}+{\cos^2x-\sin^2x\over\cos^2y-\sin^2y} \\[10pt]
= {} &
{3\sin y\cos y\over2\sin y\cos y}+{(1/4)\cos^2y-9\sin^2y\over\cos^2y-\sin^2y} \\[10pt]
= {} &
{3\over2}+{(1/4)\cos^2y-9\sin^2y\over\cos^2y-\sin^2y}.
\end{align}
On the other hand:
$$
1=\cos^2x+\sin^2x={1\over4}\cos^2y+9(1-\cos^2y),
\quad\hbox{whence:}\quad
\cos^2y={32\over35},\quad\sin^2y={3\over35}.
$$
We finally have
$$
{\sin2x\over\sin2y}+{\cos2x\over\cos2y}=
{3\over2}+{{1\over4}{32\over35}-9{3\over35}\over{32\over35}-{3\over35}}={49\over58}.
$$
