Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If: $$\frac{\cos x}{\cos y}=\frac{1}{2}$$ and $$\frac{\sin x}{\sin y}=3$$

How to find $$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}$$

share|improve this question

3 Answers 3

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.

share|improve this answer

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

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

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

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.

share|improve this answer
1  
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, $$\frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}=\frac 3 2-\frac{19}{29}=\frac{49}{58}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.