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

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. – Gerry Myerson Nov 8 '12 at 1:45
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. – DavidSalib Nov 8 '12 at 1:46

## 2 Answers

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? – DavidSalib Nov 8 '12 at 1:51
You should divide the first of Avatar's equations by the second. – Gerry Myerson Nov 8 '12 at 1:53
Therefore, sin(x+y)/sin(x-y) ? – DavidSalib Nov 8 '12 at 1:54
Yes, David; and you know that quotient is 2. Now, what happens on the other side of the equation? – Gerry Myerson Nov 9 '12 at 0:11

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 $$\frac{\tan x}{\tan y} = 3$$

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