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Can we solve the system of equations:

$$\sin \alpha + \sin \beta + \sin \gamma = 0$$

$$\cos \alpha + \cos \beta + \cos \gamma = 0$$


(i.e. find the possible values of $\alpha, \beta, \gamma$)

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up vote 11 down vote accepted

Try something similar to what has been posted yet. Take one variable to the other side, then square and add the equations. What you get is $\alpha-\beta=120°$ and same for cyclic permutations (or negating all angles). The solutions is the three angles $\delta$, $\delta+120°$, $\delta-120°$ (arbitrary $\delta$) in any order.

EDIT: Or simply realize that the equations are equivalent to $\exp(i\alpha)+\exp(i\beta)+\exp(i\gamma)=0$ which make the answer immediately obvious.

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Developing on Gerenuk's answer, you could consider the complex numbers

$$ z_1=\cos \alpha+i\sin \alpha,\ z_2=\cos \beta+i\sin\beta,\ z_3=\cos \gamma+i\sin \gamma$$

Then you know that $z_1,z_2,z_3$ are on the unit circle, and the centroid of the triangle formed by the points of afixes $z_i$ is of afix $\frac{z_1+z_2+z_3}{3}=0$. From classical geometry, we can see that if the centroid of a triangle is the same as the center of the circumscribed circle, then the triangle is equilateral. This proves that $\alpha,\beta,\gamma$ are of the form $\theta,\theta+\frac{2\pi}{3},\theta+\frac{4\pi}{3}$.

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Try writing $$ \sin^2(\alpha)=(\sin(\beta)+\sin(\gamma))^2=\sin^2(\beta)+\sin^2(\gamma)+2\sin(\beta)\sin(\gamma)\tag{1} $$ and $$ \cos^2(\alpha)=(\cos(\beta)+\cos(\gamma))^2=\cos^2(\beta)+\cos^2(\gamma)+2\cos(\beta)\cos(\gamma)\tag{2} $$ then add $(1)$ and $(2)$ to get $$ 1=1+1+2\cos(\beta-\gamma)\tag{3} $$ which means $\cos(\beta-\gamma)=-\frac12$. The same is true for the other pairs, so we get that each of $\alpha$, $\beta$, and $\gamma$ differ from each other by $\frac{2\pi}{3}$, which is the same answer that was achieved using complex means already.

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Nice Answer without complex numbers. – user21436 Feb 26 '12 at 15:25

Here's an algebraic proof : Write $z_k = e^{i \alpha_k}$. Then your equations are equivalent to $z_1 + z_2 + z_3 = 0$. Write $\theta = \frac{\alpha_1 + \alpha_2 + \alpha_3}{3}$ and $z = e^{i \theta}$

Expand the polynomial $P = (X-z_1)(X-z_2)(X-z_3)$. The $X^2$ term is $0$ by hypothetis, and the $X$ term can be written as $z_1 z_2 z_3 (z_1^* + z_2^* + z_3^*) = 0$. So $P = X^3 - z_1 z_2 z_3 = X^3 - z^3$. So the roots are $z . e^{2i k \pi /3}$.

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More of a comment and less of an answer!

Well, your information seems to tell us that,


(To see this, square both equalities and add.)

I don't see any other obvious thing, you can do with these equations. If any thought plops up, I will type it in here.

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Your relation should be with $\alpha-\beta,\beta-\gamma$ and $\gamma-\alpha$ – Beni Bogosel Feb 23 '12 at 17:58
@BeniBogosel I am sorry for having made that blunder. Thanks for the pointer. That the answer has posted, I'll flag it for the moderator to make it CW. – user21436 Feb 24 '12 at 8:01
Your result follows from my answer which shows that $$\cos(\alpha-\beta)=\cos(\beta-\gamma)=\cos(\gamma-\alpha)=-\frac12$$ – robjohn Feb 26 '12 at 23:31

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