# 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 $$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$$

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## 3 Answers

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.

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

$$\frac{\cos{2x}}{\cos{2y}}=\frac{\cos^{2}{x}-\sin^{2}{x}}{\cos^{2}{y}-\sin^{2}{y}}. \tag{2}$$

$$\frac{\cos{x}}{\cos{y}}=\frac{1}{2} \Leftrightarrow 4 \cdot\cos^{2}{x}=\cos^{2}{y}.\tag{3}$$ $$\frac{\sin{x}}{\sin{y}}=3 \Leftrightarrow \frac{1}{9}\cdot \sin^{2}{x}=\sin^{2}{y}.\tag{4}$$

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.

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Some $\iff$ should be replaced by $\implies$. – Did Sep 22 '12 at 11:37

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, $$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}=\frac 3 2-\frac{19}{29}=\frac{49}{58}$$

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