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I worked my way to encounter a system of equations

\begin{equation} \begin{cases} q=k_1\cos\phi_1+k_2\cos\phi_2& \\ 0=k_1\sin\phi_1+k_2\sin\phi_2& \end{cases} \end{equation}

I don't need to solve for specific angles $\phi_1,\phi_2$ because these equations will probably be very rough, but it's enough to know the answer in terms of $\cos\phi_1$ and $\sin\phi_2$ or vice versa. If I am looking to solve this by hand then what I have to do is obviously rewrite one of the trigonometric functions in terms of the other, for example

\begin{equation} \begin{cases} q=k_1\cos\phi_1\pm k_2\sqrt{1-\sin^2\phi_2}& \\ 0=\pm k_1\sqrt{1-\cos^2\phi_1}+k_2\sin\phi_2& \end{cases}, \end{equation} where now it's just basic algebra to solve the equations.

My question concerns the $\pm$ signs that appear there. I can rearrange the equations in such a way, that after squaring them I get a sign independent result again. Hence I get a unique solution to this equation. But to me that seems weird - I would assume I should get solutions that aren't unique. I assume that I must at some point arrive at some expression that differs by (at least) a sign, since I am expressing everything in terms of $\cos\phi_1$ and $\sin\phi_2$, which can swap signs depending on their argument. Could anyone clear this up for me?

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Let me reframe this for a bit...also let's assume $\phi_1$ and $\phi_2$ are "interesting" (not $0, \pi$, etc).

Just to be on the same page, let's solve the equation first. Isolating the square roots in both equations gives $q - k_1 \cos\phi_1 = \pm k_2 \sqrt{1-\sin^2 \phi_2}$ and $\mp k_1 \sqrt{1 - \cos^2 \phi_1} = k_2^2 \sin^2 \phi_2$. Squaring and solving gives

$$ \cos \phi_1 = \frac{q^2 + k_1^2 - k_2^2}{2qk_1} $$

which is indeed sign independent. But all of the terms with $\sin \phi_2$ are squared, so $\sin \phi_2$ is not sign independent. Solving the other way, that is, writing $q - k_2 \cos\phi_2 = \pm k_1 \sqrt{1-\sin^2 \phi_1}$ and $\mp k_2 \sqrt{1 - \cos^2 \phi_2} = k_1^2 \sin^2 \phi_1$, gives

$$ \cos \phi_2 = \frac{q^2 + k_2^2 - k_1^2}{2qk_2} $$

(which also makes sense because of symmetry). Similarly, you can't get a value for $\sin \phi_1$ here, since all the $\sin \phi_1$ terms are squared.

The first time, $\cos \phi_1$ was sign-independent and $\sin \phi_2$ wasn't. The second time, $\cos \phi_2$ was sign independent and $\sin \phi_1$ wasn't. Specifically, if you try to solve for the $\sin$s, you get expressions like $\sin \phi = \pm \text{(stuff)}$.

Think of the unit circle for a second. Geometrically, we know the x-coordinates of $\phi_1$ and $\phi_2$, but not their y-coordinates. Though we do know the two possible candidates are $\pm$ of each other. So we are almost done but not quite. We have four possible solutions: $(\phi_1, \phi_2)$, $(\phi_1, -\phi_2)$, $(-\phi_1, \phi_2)$, $(-\phi_1, -\phi_2)$.

Note that $(\phi_1, \phi_2)$ if a solution if and only if $(-\phi_1, -\phi_2)$ is, and the same holds for the two other solution candidates $(\phi_1, -\phi_2)$, $(-\phi_1, \phi_2)$. (See this by plugging them into the second equation.) And $(\phi_1, \phi_2)$ and $(\phi_1, -\phi_2)$ can't both be solutions because the second equation would give $\sin \phi_1 = 0$, a contradiction. Since $(\phi_1, \phi_2)$ is of course a solution to the equation, this tells you that the only two solutions of the equation are $(\phi_1, \phi_2)$ and $(-\phi_1, -\phi_2)$.

This is a long-winded way of saying: You do get solutions that aren't unique when solving this. Specifically, you get 4 solutions, and 2 of them turn out to be fake solutions, leaving you with 2 actual ones. I can't tell where exactly in your question you went wrong, because you didn't explain your solution path, but hopefully this clears things up.

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  • $\begingroup$ Yeah, your explanation makes sense. I didn't give too much information, since I actually wanted to solve this one on my own (it's not that hard after all). I'll try to apply this to my problem. Thanks. $\endgroup$ – Henrikas Aug 16 '18 at 8:13

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