Simultaneous equations, trig functions and the existence of solutions Came across this conundrum while going over the proof that
$$A \cdot \sin(bx) + B \cdot \cos(bx) = C \cdot \sin(bx + k)$$
for some numbers $C$ and $k$. ($A$, $B$ and $b$ are known.)
The usual method is to expand the RHS using the compound angle identity
\begin{align}
C \cdot \sin(bx + k) &= C \cdot \bigl( \sin(bx)\cos(k) + \cos(bx)\sin(k) \bigl) \\
&= C\cos(k) \cdot \sin(bx) + C\sin(k) \cdot \cos(bx)
\end{align}
and thus set
\begin{align}
C\cos(k) &= A \\
C\sin(k) &= B
\end{align}
My trouble comes with what happens at this point - we then proceed to divide the second equation by the first, obtaining
$$ \tan(k) = \frac{B}{A} $$
which we then solve to obtain $k$, etc. etc.
My question is: how do we know that this is "legal"? We have reduced the original two-equation system to a single equation. How do we know that the values of $k$ that satisfy the single equation are equal to the solution set of the original system?
While thinking about this I drew up this other problem:
\begin{align}
\text{Find all }x\text{ such that} \\
\sin(x) &= 1 \\
\cos(x) &= 1
\end{align}
Obviously this system has no solutions ($\sin x$ and $\cos x$ are never equal to $1$ simultaneously). But if we apply the same method we did for the earlier example, we can say that since $\sin(x) = 1$ and $\cos(x) = 1$, let's divide $1$ by $1$ and get
$$ \tan(x) = 1 $$
which does have solutions.
So how do we know when it's safe to divide simultaneous equations by each other? (If ever?)

In general, you don't. You know that all of the solutions of the pair of equations you started with are solutions of the single equation you ended up with (barring division-by-zero issues), but you generally don't know the converse. In this case, the reason you can get away with the converse is that you can choose $C$. Knowing $\tan k$ is the same as knowing $(\cos k, \sin k)$ up to a multiplicative constant; draw a unit circle if you don't believe this.
In general, the only way you know whether it is "legal" to do anything is to prove or disprove that you can do it.
