If $\sin \alpha+\sin \beta+\sin \gamma=0$ and $\cos \alpha+\cos \beta+\cos \gamma=0$
Then find the values of $\sum \sin^2 \alpha$ and $\sum \cos^2 \alpha$.
Try: $$(\sin \alpha+\sin \beta+\sin \gamma)^2=\sum \sin^2 \alpha+2\sum \sin (\alpha+\beta)$$
So $$\sum\sin^2 \alpha = -2\sum \sin (\alpha+\beta)$$
Similarly $$\sum \cos^2 \alpha = -2\sum \cos(\alpha-\beta)$$
From $\sin \alpha+\sin \beta = -\sin \gamma$ and $\cos \alpha+\cos \beta = -\cos \gamma$
Squaring and Adding $$2+2\cos(\alpha-\beta)=1\Rightarrow \cos (\alpha-\beta)=-\frac{1}{2}$$
$$\sum \cos^2 \alpha =3 = \sum \sin^2 \alpha$$
Could someone explain me in geometrical way.
I have seems that
$(\cos \alpha,\sin \alpha)\;\;,(\cos \beta,\sin \beta)\;\;,(\cos \gamma,\sin \gamma)$ must be the vertices of an triangle.
But not understand how to solve further thanks
Best Answer
Let $A(\cos\alpha,\sin\alpha)$, $B(\cos\beta,\sin\beta)$ and $C(\cos\gamma,\sin\gamma).$
Thus, $OA=OB=OC=1$ and since $$\vec{OA}+\vec{OB}+\vec{OC}=\vec{0},$$ we obtain that $O$ is a circumcenter and a center of gravity of $\Delta ABC.$
Thus, our triangle is an equilateral triangle and we obtain:
$$\measuredangle AOB=\measuredangle AOC=\measuredangle BOC=120^{\circ},$$ which gives $$\cos(\alpha-\beta)=\cos(\alpha-\gamma)=\cos(\beta-\gamma)=-\frac{1}{2}$$ and the rest is smooth: $$\sum_{cyc}\sin^2\alpha=\left(\sum\limits_{cyc}\sin\alpha\right)^2-2\sum_{cyc}\sin\alpha\sin\beta=$$ $$=\sum_{cyc}(\cos(\alpha+\beta)-\cos(\alpha-\beta))=-\sum_{cyc}(\cos\gamma+\cos(\alpha+\beta))=\frac{3}{2}.$$ By the same way we can get $$\sum_{cyc}\cos^2\alpha=\frac{3}{2}.$$