I'm trying to solve the following system:
$$K \sin(s-t) = \sin(s)$$
$$K \sin(t-s) = \sin(t)$$
I've tried playing around with it and learned that $\sin(t) = - \sin(s)$, so $s = -t + 2 n \pi$ for all $n \in \mathbb{N}$. However, that's not a solution. What I'm really look for is $s$ and $t$ in terms of $K$.
Mathematica gives a solution involving $\arctan$, however I don't know how to come up with that on my own and I would like to.
Is there some inverse trig trick that I'm missing?
Hint:
From $\sin t=\sin(-s)\Rightarrow -s=t+2k\pi$ or $-s=-t+(2k+1)\pi$. Now you consider each case separately.
1) $-s=t+2k\pi\Rightarrow$ plug it in, say the second equation: $K\sin(t-s)=\sin t$ $$K\sin(t+t+2k\pi)=\sin t\Leftrightarrow K\sin(2t+2k\pi)=\sin t\Leftrightarrow K\sin 2t=\sin t\Leftrightarrow 2K\sin t\cos t=\sin t$$ Now consider again two cases: $\sin t=0\,$ or $\,\sin t\neq 0$. If $\sin t=0$, see the second case. If not, divide by $\sin t\neq 0$ and get $2K\cos t=1\Rightarrow \cos t=\frac{1}{2K}$. For this we obviously need $\frac{1}{2K}\in[-1,1]$. Then $t=\arccos\frac{1}{2K}$ where the usual principle value for $t$ is in $[0,\pi]$.
2) $-s=-t+(2k+1)\pi$. Again plug it in, say the second equation: $$K\sin(t-t+(2k+1)\pi)=\sin t\Leftrightarrow K\sin ((2k+1)\pi)=\sin t\Leftrightarrow 0=\sin t\Rightarrow t=l\pi,\,l\in\mathbb Z$$ and the equation is satisfied no matter what is $K$. Therefore the solution of the system is $t=l\pi,\,l\in\mathbb Z,\,-s=t=2k\pi=l\pi+2k\pi,\,k\in\mathbb Z,\,\forall K\in\mathbb R$