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How can I find $x,y$ for:

$$\sin x + \sin y = 0 \\ \sin 2x + \sin 2y = 0$$

I've tried to utilize the identity $\sin 2x = 2\sin x\cos x$ but wasn't able to reach the solution.

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    Hint: there is not one unique solution. – 2017-02-27
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    the solution to the first line is. $y = n\pi - (-1)^n x$ the 2nd line will be similar. – 2017-02-27

3 Answers 3

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You've used the correct identity, just go on:

$$\sin \, x = - \sin \,y$$

hence the second equations is equivalent to

$$\sin y (\cos \,y - \cos \, x) = 0$$ and now you just split the solution into

$$\sin \,y = 0\ \text{ OR }\ (\cos \,y - \cos \, x) = 0.$$

In the first case, when $\sin \,y = 0$, then $\sin \,x = 0$ as well, so $(x,y) = (k;n)\pi$ for $k,n \in \mathbb{N}$.

In the second case one has to consider both conditions, i.e. $\cos \,y = \cos \, x$ implies $x$ and $y$ must be either equal or $x=-y$ (symmetrical along horizontal axis + consider the period) from which the condition $\sin \, x = - \sin \,y$ chooses the latter.

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    so basically any $x=y$ will do, right? – 2017-02-27
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    @pepa.dvorak Wait, surely `any x = y will do` is incorrect. I can definitely name a value that fits `x = y` but makes the first equation false. – 2017-02-27
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HINT:

$$\sin z=-\sin A=\sin(-A)$$

$$\implies z=n\pi+(-1)^n(-A)$$ where $n$ is any integer

Put $z=x, A=y$ and then $z=2x,A=2y$

Find which values of $x$ satisfy both the equations.

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HINT: your System is equivalent to $$2\cos\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right)=0$$ and $$2\cos(x-y)\sin(x+y)=0$$ can you proceed?

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    http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html – 2017-02-27