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?)