<s>[INST] 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?
 [/INST] \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
$$\frac{8\pi^2}{9}$$
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