# Solving the system $\sum \sin = \sum \cos = 0$.

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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## 5 Answers

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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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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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

More of a comment and less of an answer!

Well, your information seems to tell us that,

$$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\alpha-\gamma)=-\dfrac{3}{2}$$

(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

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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