If $\sin\alpha+\sin\beta+\sin\gamma=\cos\alpha+\cos\beta+\cos\gamma=0$, then find the values of $\sum \sin^2 \alpha$ and $\sum \cos^2 \alpha$.

Question

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

Answer 1

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 $$\sum_{cyc}\cos^2\alpha=\frac{3}{2}.$$