Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$ Find $x^2+y^2+z^2$ if $x+y+z=0$ and $\sin{x}+\sin{y}+\sin{z}=0$ and $\cos{x}+\cos{y}+\cos{z}=0$ for any $x,y,z \in [-\pi,\pi]$.
My attempt:I found one answer $x=0,y=\frac{2\pi}{3},z=-\frac{2\pi}{3}$ which gives the answer $x^2+y^2+z^2=\frac{8{\pi}^2}{9}$.But I want a way to find the answer using equations.ANy hints?
 A: You can try this method:
$$x+y+z=0 \implies x+y=-z\tag1$$
And $$\sin{x}+\sin{y}+\sin{z}=0$$
$$\implies \sin{x}+\sin{y}=-\sin{z}\tag2$$
And $$\cos{x}+\cos{y}+\cos{z}=0$$
$$\cos{x}+\cos{y}=-\cos{z}\tag3$$
So $$(2)^2+(3)^2 \implies 2+2\cos(x-y)=1$$
$$\implies \cos(x-y)=-\frac12\tag4$$
Similarly we get $$\cos(y-z)=-\frac12\tag5$$
$$\cos(z-x)=-\frac12\tag6$$
And from $(1)$, we get $\cos(x+y)=\cos z$
Can you proceed now?
A: \begin{eqnarray}
\sin z &=& -\sin x-\sin y\\
\cos z &=& -\cos x-\cos y
\end{eqnarray}
so
\begin{eqnarray}
\sin^2z &=& \sin^2x+\sin^2y+2\sin x\sin y\\
\cos^2z &=& \cos^2x+\cos^2y+2\cos x\cos y
\end{eqnarray}
adding to these equations concludes
\begin{eqnarray}
1 &=& 1+1+2(\sin x\sin y+\cos x\cos y)\\
0 &=& 1+2\cos(x-y)\\
-\frac{1}{2} &=& \cos(x-y)
\end{eqnarray}
and also
$$\cos^2\frac{x-y}{2}=\frac{1+\cos(x-y)}{2}=\frac{1}{4}$$
or
$$\cos\frac{x-y}{2}=\pm\frac{1}{2}$$
In the other hand
\begin{eqnarray}
(\sin x+\sin y+\sin z)^2 + (\cos x+\cos y+\cos z)^2 &=& 0\\
\sin^2x+\sin^2y+\sin^2z+2\sin x\sin y+2\sin y\sin z+2\sin z\sin x &+&\\
\cos^2x+\cos^2y+\cos^2z+2\cos x\cos y+2\cos y\cos z+2\cos z\cos x &=&0\\
3+2\big(\cos x\cos y+\sin x\sin y\big)+2\big(\cos y\cos z+\sin y\sin z\big)+2\big(\cos z\cos x+\sin z\sin x\big) &=&0\\
3+2\cos(x-y)+2\cos(y-z)+2\cos(z-x) &=&0\\
3+2(-\frac{1}{2})+4\cos\frac{x-y}{2}\cos3\frac{x+y}{2} &=&0\\
\cos\frac{x-y}{2}\cos3\frac{x+y}{2} &=&-\frac{1}{2}\\
\cos3\frac{x+y}{2} &=&\mp1\\
\end{eqnarray}
so
\begin{eqnarray}
\cos(x-y) &=& -\frac{1}{2}\\
\cos3\frac{x+y}{2} &=&\mp1
\end{eqnarray}
We have two answer for each of them
\begin{eqnarray}
x-y &=& \frac{2\pi}{3},~\frac{-\pi}{3}\\
x+y &=& 0,~\frac{2\pi}{3}
\end{eqnarray}
We obtain 4 answer from these equations and then permutation over $(x,y,z)$ from equations symmetrically, gives us 12 answer. Finally by substitution answers in equations, $x^2+y^2+z^2$ has one value ($\displaystyle\frac{2\pi}{3},\frac{-2\pi}{3},0$), it is
$$\frac{8\pi^2}{9}$$
A: Although this is same to answer of @schrodingersCat　Let $x\geq y\geq0\geq z$ 
$$\cos(x+y)^2=1-\sin(x+y)^2$$ 
$$⇔1-(\sin{x}+\sin{y})^2=(\cos{x}+\cos{y})^2$$
$$⇔1+2(\cos{x}\cos{y}+\sin{x}\sin{y})=0$$
$$⇔\cos(x-y)=-1/2 $$
$$x-y=2\pi/3$$
and
$$\sin{x}+\sin{y}=-\sin{z}$$
$$⇔\sin(y+2\pi/3)+\sin{y}=\sqrt3/2$$ 
$$⇔(1/2)\sin{y}+\sqrt3/2\cos{y}=\sqrt3/2$$
$$⇔\sin(y+\pi/3)=\sqrt3/2$$ 
then we get $$x=2\pi/3 ,y=0,z=-2\pi/3$$, if angle(x,y,z) are points of unit circle, they consist of the regular triangle. The answer of $x^2+y^2+z^2$  is unique and,
$$\displaystyle x^2+y^2+z^2=\frac{8\pi^2}9 $$
A: HINT.- $$\begin{cases}\sin{x}+\sin{y}+\sin{z}=0\\\cos{x}+\cos{y}+\cos{z}=0\end{cases}\iff\begin{cases}2\sin\dfrac{x+y}{2}\cos\dfrac{x-y}{2}=-\sin z\\2\cos\dfrac{x+y}{2}\cos\dfrac{x-y}{2}=-\cos z\end{cases}\Rightarrow \tan\frac{x+y}{2}=\tan z$$
Hence, because of $x,y,z \in [-\pi,\pi]$, $$x+y=2z\iff x+y+z=3z\Rightarrow \color{red}{z=0 \text{  and }x+y=0}$$
NOTE.-Obviously instead of $z$ we can choose either $x$ and $y$ to be zero.  Thus $$\color{red}{x^2+y^2+z^2=2t^2}$$ where $(x,y,z)=(t,-t,0)$ and with $t\in [-\pi,\pi]$.
