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I encountered the following set of equations whilst trying to solvea a problem, to say I am well versed in multivariate calculus of this complexity would be a blatant lie. I hope somebody here could tell me if the following are solveable and/or point me in the direction to learn more.

$$\ \ \ \frac{\sin \left ( 2\left ( \alpha _0 -\alpha _1 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _1 \right )}+\cos^{2} \left ( \beta _0 -\beta _1 \right ) }} \ + \frac{\sin \left ( 2\left ( \alpha _0 -\alpha _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _2 \right )}+\cos^{2} \left ( \beta _0 -\beta _2 \right ) }} = 0 \\ \ \ \ \ \frac{\sin \left ( 2\left ( \beta _0 -\beta _1 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _1 \right )}+\cos^{2} \left ( \beta _0 -\beta _1 \right ) }} + \frac{\sin \left ( 2\left ( \beta _0 -\beta _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _2 \right )}+\cos^{2} \left ( \beta _0 -\beta _2 \right ) }} = 0 \\ -\frac{\sin \left ( 2\left ( \alpha _0 -\alpha _1 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _1 \right )}+\cos^{2} \left ( \beta _0 -\beta _1 \right ) }} + \frac{\sin \left ( 2\left ( \alpha _1 -\alpha _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _1 -\alpha _2 \right )}+\cos^{2} \left ( \beta _1 -\beta _2 \right ) }} = 0 \\ \ \ \ \ \frac{\sin \left ( 2\left ( \beta _0 -\beta _1 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _1 \right )}+\cos^{2} \left ( \beta _0 -\beta _1 \right ) }} - \frac{\sin \left ( 2\left ( \beta _1 -\beta _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _1 -\alpha _2 \right )}+\cos^{2} \left ( \beta _1 -\beta _2 \right ) }} = 0 \\ -\frac{\sin \left ( 2\left ( \alpha _0 -\alpha _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _2 \right )}+\cos^{2} \left ( \beta _0 -\beta _2 \right ) }} - \frac{\sin \left ( 2\left ( \alpha _1 -\alpha _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _1 -\alpha _2 \right )}+\cos^{2} \left ( \beta _1 -\beta _2 \right ) }} = 0 \\ \ \ \ \ \frac{\sin \left ( 2\left ( \beta _0 -\beta _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _2 \right )}+\cos^{2} \left ( \beta _0 -\beta _2 \right ) }} + \frac{\sin \left ( 2\left ( \beta _1 -\beta _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _1 -\alpha _2 \right )}+\cos^{2} \left ( \beta _1 -\beta _2 \right ) }} = 0 $$

My intuition of the underlying problem tells me that if it is in some sense solvable one would need to fix a pair $(\alpha_n,\beta_n) = (0,0)$ for example to get a solution set for the other pairs and fixing another one of these pairs $(\alpha_n,\beta_n)$ whilst also satisfying that solution set will give you a value for the last pair.

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    Can you add where these equations come from? – 2017-02-13
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    An immediate observation is that if $\alpha_0 = \alpha_1 = \alpha_2$ and $\beta_0 = \beta_1 = \beta_2$, then all the numerators are $0$, but all the denominators are $1$. So this will be a solution. – 2017-02-13

1 Answers 1

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Your intuition is right: if you have one solution consisting of the three alphas and the three betas, and you add the same number to all three alphas. and some different number to all three betas, you'll still have a solution. So your problem is somewhat redundant: despite having 6 vars and six equations, rather than there being a single solution, you're in a situation where if there's even one solution, then there's a whole PLANE of solutions. Indeed, by @Alfred Yerger's comment, the plane defined by $\alpha_0 = \alpha_1 = \alpha_2, \beta_0 = \beta_1 = \beta_2$ is one possible plane of solutions.

As for the equations themselves...you can make things look a bit nicer at the risk of introducing extraneous solutions, by squaring some things. For instance, the first equationn can be shuffled across the equals sign and then squared:

\begin{align} \frac{\sin \left ( 2\left ( \alpha _0 -\alpha _1 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _1 \right )}+\cos^{2} \left ( \beta _0 -\beta _1 \right ) }} \ + \frac{\sin \left ( 2\left ( \alpha _0 -\alpha _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _2 \right )}+\cos^{2} \left ( \beta _0 -\beta _2 \right ) }} = 0\\ \frac{\sin \left ( 2\left ( \alpha _0 -\alpha _1 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _1 \right )}+\cos^{2} \left ( \beta _0 -\beta _1 \right ) }} \ = -\frac{\sin \left ( 2\left ( \alpha _0 -\alpha _2 \right ) \right )}{2\sqrt{\sin^{2} {\left ( \alpha _0 -\alpha _2 \right )}+\cos^{2} \left ( \beta _0 -\beta _2 \right ) }} \\ \frac{\sin^2 \left ( 2\left ( \alpha _0 -\alpha _1 \right ) \right )}{4\left({\sin^{2} {\left ( \alpha _0 -\alpha _1 \right )}+\cos^{2} \left ( \beta _0 -\beta _1 \right ) } \right)} \ = \frac{\sin^2 \left ( 2\left ( \alpha _0 -\alpha _2 \right ) \right )}{4 \left( {\sin^{2} {\left ( \alpha _0 -\alpha _2 \right )}+\cos^{2} \left ( \beta _0 -\beta _2 \right ) }\right)}\\ \end{align} Now you can turn $\sin (2 (a_0 - a_1 ))$ into $ 2 \sin(a_0 - a_1) \cos(a_0 - a_1)$, which you then square, and do the same on the right hand numerator. That'll give you factors of 4 that cancel this in the denominator: \begin{align} \frac{ \left(\sin ( \alpha _0 -\alpha_1 ) \cos(\alpha_0 - \alpha_1) \right)^2}{{\sin^{2} {\left ( \alpha _0 -\alpha _1 \right )}+\cos^{2} \left ( \beta _0 -\beta _1 \right ) } } \ = \frac{\left(\sin ( \alpha _0 -\alpha_2 ) \cos(\alpha_0 - \alpha_2) \right)^2}{ {\sin^{2} {\left ( \alpha _0 -\alpha _2 \right )}+\cos^{2} \left ( \beta _0 -\beta _2 \right ) }} \end{align}

Now you can cross-multiply, and gather some like terms, and I think that things will get at least a little more manageable. You might want to remember the sine and cosine addition and subtraction formulas as you do so.