<s>[INST] $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=\frac{-3}{2}$,show that $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$ I think that I've done a major part of the problem but I'm stuck at a point.

Here's what I've done :
It's given to us that
$$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha) = \dfrac{-3}{2}$$
Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we obtain :
$$\cos\alpha\cos\beta + \sin\alpha\sin\beta + \cos\beta\cos\gamma + \sin\beta\sin\gamma + \cos\gamma\cos\alpha + \sin\gamma\sin\alpha = \dfrac{-3}{2}$$
Multiplying both sides by $2$, we obtain :
$$2\cos\alpha\cos\beta + 2\cos\beta\cos\gamma + + 2\cos\gamma\cos\alpha + 2\sin\alpha\sin\beta + 2\sin\beta\sin\gamma + 2\sin\gamma\sin\alpha = -3$$
Adding $\sin^2\alpha+\sin^2\beta+\sin^2\gamma+\cos^2\alpha+\cos^2\beta+\cos^2\gamma$ to both sides, we obtain :
$$\text{LHS : } (\cos^2\alpha + \cos^2\beta + \cos^2\gamma + 2\cos\alpha\cos\beta + 2\cos\beta\cos\gamma + 2\cos\gamma\cos\alpha)$$
$$ + (\sin^2\alpha + \sin^2\beta + \sin^2\gamma + 2\sin\alpha\sin\beta + 2\sin\beta\sin\gamma + 2\sin\gamma\sin\alpha)$$
$$\text{RHS : } -3 + (\cos^2\alpha + \sin^2\alpha) + (\cos^2\beta + \sin^2\beta) + (\cos^2\gamma + \sin^2\gamma)$$
On simplifying,
$$\text {LHS : } (\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma)^2$$
$$\text{RHS : } -3+1+1+1 = -3+3 = 0$$
So, we obtain :
$$(\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma)^2 = 0$$
$$\implies (\cos\alpha + \cos\beta + \cos\gamma)^2 = -(\sin\alpha + \sin\beta + \sin\gamma)^2$$
Now, square rooting both sides would involve $\iota$ i.e. $\sqrt{-1}$ but I haven't learnt about complex numbers yet and I think that the solution can be continued without using complex numbers but I don't know how.
Any help would be appreciated.

Thanks!
 [/INST] If the sum of squares of two real numbers is zero, it implies that both numbers are zero. If you want you can simply prove this using Reductio-Ad-Absurdum.
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