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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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    David, all kidding aside, you're question will get a better reception if you let us know why it interests you, what you know about it, what ideas you might know that are relevant to solving it, where you got stuck, and so on.2012-11-08
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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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    Ok, I see what you did... should I bring the right side over to the left side OR simplify each side if possible?2012-11-08
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    You should divide the first of Avatar's equations by the second.2012-11-08
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    Therefore, sin(x+y)/sin(x-y) ?2012-11-08
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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 $$ \frac{\tan x}{\tan y} = 3 $$