Rewriting $\sin(a+b) = c$ Given $$\sin(a+b) = c$$
Could it be rewritten as $$a = \arcsin(c) - b$$
For all reals $a$ and $b$?
Sine over reals isn't one-to-one though. Is the above valid?

In general $\sin(x)=y$ is equivalent to
$$ x=\pi n+(-1)^n\arcsin(y)\text{ for some }n\in\mathbb{Z} $$
So if $\sin(a+b)=c$ then there exists an integer $n$ such that
\begin{equation} a=\pi n+(-1)^n\arcsin(c) -b \tag{1}   \end{equation}
This is equivalent to the WolframAlpha solution but is more compact.
Note that
\begin{eqnarray}
\sin(\pi n+(-1)^n\arcsin(y))&=&\sin(\pi n)\cos((-1)^n\arcsin(y))\\
& &+\cos(\pi n)\sin((-1)^n\arcsin(y))\\
&=&0+(-1)^n(-1)^n\sin(\arcsin(y))\\
&=&y
\end{eqnarray}
