Sign up ×
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 $2\sin(x-y) = \sin(x+y)$, 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.

share|cite|improve this question
Why is this urgent? – Jason DeVito Nov 8 '12 at 1:39
Should we answer it, perhaps it prevents the next album of Justin Bieber to be released$\ldots$ – Jean-Sébastien Nov 8 '12 at 1:43
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
@DavidSalib Welcome to math.SE. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, kindly avoid phrases like "Prove this", "Show that", "Urgent" etc. Also, avoid capital letters since people interpret this as shouting at someone. If it is a homework, kindly tag it as homework. – user17762 Nov 8 '12 at 1:48
Ok thank you @Marvis – DavidSalib Nov 8 '12 at 1:49

2 Answers 2

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

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

share|cite|improve this answer
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 $$ \frac{\tan x}{\tan y} = 3 $$

share|cite|improve this answer

Your Answer


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.