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If $2\sin(x-y) = \sin(x+y)$, find $\displaystyle \frac{\tan(x)}{\tan(y)}$

= 2(sinxcosy-sinycosx) - sinxcosy - sinycosx = sinxcosy - sinycosx

//I'm not sure if bringing the sin(x + y) to the left side was right or wrong, just an idea I had.

Thanks in advance for your help, very appreciated.

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    You're all hilarious... However it's actually a homework question that is likely on the my test tomorrow worth 30% of the mark. I'll edit to show some steps.2012-11-08

2 Answers 2

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Hint: Use $\sin(x+y)=\sin x\cos y+\cos x\sin y=\cos x\cos y(\tan x+\tan y)$ and

$\sin(x-y)=\sin x\cos y-\cos x\sin y=\cos x\cos y(\tan x-\tan y)$

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    Yes, David; and you know that quotient is 2. Now, what happens on the other side of the equation?2012-11-09
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Now that the question is no longer homework, I can expand on Avatar's solution. Assuming $\cos x\text{ and }\cos y$ aren't zero (otherwise the tangents are undefined), we'll have $ \begin{align} 2\sin(x-y) &= \sin(x+y) & \text{so, by definition}\\ 2(\sin x \cos y-\cos x\sin y) &= \sin x\cos y + \cos x\sin y & \text{so}\\ 2\sin x \cos y-2\cos x\sin y &= \sin x\cos y + \cos x\sin y & \text{collect terms to get}\\ \sin x\cos y &= 3\cos x\sin y & \text{divide by $\cos x\cos y$}\\ \tan x &= 3\tan y \end{align} $ so $ \frac{\tan x}{\tan y} = 3 $