finding $\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$ If:
$$\frac{\cos x}{\cos y}=\frac{1}{2}$$  and $$\frac{\sin x}{\sin y}=3$$
How to find 
$$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$$
 A: Hint: First multiply the two equations to get $\frac{\sin 2x}{\sin 2y}=\frac{3}{2}$.
Now, squaring the first equation to get $\cos^2x=\frac{\cos^2y}{4}$ and similarly from the second $\sin^2x=9\sin^2y$. Adding these two, you will get the value of $\cos^2y$ and substituting into anyone of these will yield the value of $\cos^2x$. Next use the formula $\cos2A=2\cos^2A-1$ to get the other term.
A: $$\frac{\sin{2x}}{\sin{2y}}=\frac{2\sin{x}\cos{x}}{2\sin{y}\cos{y}}=\frac{\sin{x}}{\sin{y}}\cdot \frac{\cos{x}}{\cos{y}}=3\cdot \frac{1}{2}=\frac{3}{2}. \tag{1}$$
$$\frac{\cos{2x}}{\cos{2y}}=\frac{\cos^{2}{x}-\sin^{2}{x}}{\cos^{2}{y}-\sin^{2}{y}}. \tag{2}$$
$$\frac{\cos{x}}{\cos{y}}=\frac{1}{2} \Leftrightarrow 4 \cdot\cos^{2}{x}=\cos^{2}{y}.\tag{3}$$
$$\frac{\sin{x}}{\sin{y}}=3 \Leftrightarrow \frac{1}{9}\cdot \sin^{2}{x}=\sin^{2}{y}.\tag{4}$$
We know that (using $(3)$ and $(4)$): 
\begin{cases}
\sin^{2}{x}+\cos^{2}{x}=1\\
4\cdot \cos^{2}{x}+\frac{1}{9}\cdot\sin^{2}{x}=1
\end{cases}
So: $\displaystyle 35\cdot\cos^{2}{x}=8 \Rightarrow \cos^{2}{x}=\frac{8}{35}\tag{5}$ and $\displaystyle \sin^{2}{x}=1-\frac{8}{35}=\frac{27}{35}. \tag{6}$
But using $(3)$ and $(4)$ we obtain that: 
$$\sin^{2}{y}=\frac{3}{35}\tag{7}$$ and $$\cos^{2}{y}=\frac{32}{35}\tag{8}$$
So $(3)$ is equivalent with : 
$$\large\frac{\frac{8}{35}-\frac{27}{35}}{\frac{32}{35}-\frac{3}{35}}=\frac{-\frac{19}{35}}{\frac{29}{35}}=-\frac{19}{29}.\tag{9}$$
The final answer is obtained using $(1)$ and $(9)$:  
$$\frac{3}{2}-\frac{19}{29}=\frac{49}{58}.$$
I hope it is all right, I hope not to mistake to calculations. 
A: Let $$\frac {\cos x}{1}=\frac {\cos y}{2}=a(say),\implies \cos x=a,\cos y =2a$$
and $$\frac{\sin x }{3}=\frac{\sin y }{1}=b(say),\implies \sin x=3b, \sin y =b$$
So, $$a^2+(3b)^2=1,(2a)^2+b^2=1\implies a^2=\frac 8{35}, b^2=\frac 3{35}$$
$$\implies \cos^2x=a^2=\frac 8{35},\sin^2y=b^2=\frac 3{35}$$
So, $$\frac{\sin 2x}{\sin 2y}=\frac{2\sin x\cos x}{2\sin y \cos y}=\frac{3b\cdot a}{b\cdot 2b}=\frac 3 2$$ as $ab \neq 0$ and
$$\frac{\cos 2x}{\cos 2y}=\frac{2\cos^2x-1}{1-2\sin^2y}=\frac{2\frac 8{35}-1}{1-2\frac 3{35}}=-\frac{19}{29}$$
So,  $$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}=\frac 3 2-\frac{19}{29}=\frac{49}{58}$$
