# If $\frac{\sin x}{\sin y}= {3}$ and $\frac{\cos x}{\cos y}= \frac{1}{2}$, then find $\frac{\sin2x}{\sin2y}+\frac{\cos2x}{\cos2y}$

Let $x$ and $y$ be real numbers such that $\dfrac{\sin x}{\sin y}= {3}$ and $\dfrac{\cos x}{\cos y}= \dfrac{1}{2}$. Find a value of $$\dfrac{\sin2x}{\sin2y}+\dfrac{\cos2x}{\cos2y}.$$

My attempts:

Using the given condition and double angle formula, $\dfrac{\sin2x}{\sin2y} = \dfrac{3}{2}$

Now I am struggling to find the value of $\dfrac{\cos2x}{\cos2y}$

The simplest form I could reach was: $\dfrac{2\cos^2x-1}{8\cos^2x-1}$

How do I continue from here? Simpler methods to solve the problem are welcome.

PS: The answer is $\dfrac{49}{58}$

• How did you get $\frac{3}{2}$? Shouldn't it be $\frac{\sin 2x}{\sin 2y} = \frac{1}{6}$? – platty Aug 2 '17 at 15:26

$\sin{y}=3\sin{x}$ and $\cos{y}=2\cos{x}$.

Thus, $$1=\sin^2y+\cos^2y=9\sin^2x+4\cos^2x=4+5\sin^2x,$$ which is impossible.

If $\sin{y}=\frac{1}{3}\sin{x}$ we obtain: $$1=\sin^2y+\cos^2y=\frac{1}{9}\sin^2x+4\cos^2x=\frac{1}{9}+\frac{35}{9}\sin^2x.$$ Now, you can get $\sin^2x$, $\sin^2y$ and the rest for you.

• @Abcd We have no contradictions in math! I think there is a mistake in the given. – Michael Rozenberg Aug 2 '17 at 15:37

hint

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

$$\implies \sin^2(x )+36\cos^2 (x)=9$$ $$\implies 35\cos^2 (x)=8$$ $$\implies \cos^2 (x)=\frac {8}{35}$$

the second expression is $$\frac {16-35}{64-35}=-19/29$$ the final result is $$3/2-19/29=49/58$$