If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then prove that $$\sin(2x-y-z)+\sin(2y-z-x)+\sin(2z-x-y)=0\\ \cos(2x-y-z)+\cos(2y-z-x)+\cos(2z-x-y)=0$$
My Attempt $$ e^{ix}+e^{iy}+e^{iz}=0=e^{-ix}+e^{-iy}+e^{-iz}\\ e^{2ix}+e^{2iy}+e^{2iz}+e^{i(x+y)}+e^{i(y+z)}+e^{i(z+x)}=0\\ e^{2ix}+e^{2iy}+e^{2iz}=0\quad\&\quad e^{i(x+y)}+e^{i(y+z)}+e^{i(z+x)}=0\\ $$
In fact I like your approachwith the exponentials.
$\begin{cases} \cos(x)+\cos(y)+\cos(z)=0\\ \sin(x)+\sin(y)+\sin(z)=0\end{cases}\implies e^{ix}+e^{iy}+e^{iz}=0$
Let's call these terms respectively $a,b,c$ we have $a+b+c=0$.
$a^3+b^3+c^3 = \underbrace{(a+b+c)^3}_{0}-3\underbrace{(a+b)}_{-c}\underbrace{(b+c)}_{-a}\underbrace{(c+a)}_{-b}=3abc$
On the other hand
$\displaystyle e^{i(2x-y-z)}+e^{i(2y-x-z)}+e^{i(2z-y-z)}=\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}=3$
Thus $\displaystyle \sum \sin(\cdots)=\Im(3)=0$
$$\sin(2x-y-z)+\sin(2y-z-x)+\sin(2z-x-y)=$$ $$=\sin(3x-x-y-z)+\sin(3y-x-y-z)+\sin(3z-x-y-z)$$ Let's take the transformation $x'=\frac{x}{3}$, $y'=\frac{y}{3}$, $z'=\frac{z}{3}$. Then, using the formula of the sine of the sum of two angles: $$\sin(3x-x-y-z)+\sin(3y-x-y-z)+\sin(3z-x-y-z)=$$ $$=\sin(x'-\tfrac{x'+y'+z'}{3})+\sin(y'-\tfrac{x'+y'+z'}{3})+\sin(z'-\tfrac{x'+y'+z'}{3})=$$ $$=\sin(x')\cos(\tfrac{x'+y'+z'}{3})-\cos(x')\sin(\tfrac{x'+y'+z'}{3})+\sin(y')\cos(\tfrac{x'+y'+z'}{3})$$ $$-\cos(y')\sin(\tfrac{x'+y'+z'}{3})+\sin(z')\cos(\tfrac{x'+y'+z'}{3})-\cos(z')\sin(\tfrac{x'+y'+z'}{3})=$$ $$\cos(\tfrac{x'+y'+z'}{3})[\sin(x')+\sin(y')+\sin(z')]-\sin(\tfrac{x'+y'+z'}{3})[\cos(x')+\cos(y')+\cos(z')]=0,$$
since $\sin(x')+\sin(y')+\sin(z')=\cos(x')+\cos(y')+\cos(z')=0$ by hypothesis.
If $\, a := \cos(x)+i\sin(x),\, b:=\cos(y)+i\sin(y),\, c:=\cos(z)+i\sin(z)\,$and if $\,a+b+c=0\,$ then $\,\Im (a^2/(bc) + b^2/(ac) +c^2/(ab))=0,\,$ since $a,b,c$ are the vertices of an equilateral triangle, and then $\,a^2/(bc)=b^2/(ac)=c^2/(ac)=1.$ A variation is that $(a^2/(bc) + b^2/(ac) +c^2/(ab)) = (a^3+b^3+c^3)/(abc)=3\,$ if $\,a+b+c=0.$
Taking $\sin(2x-y-z)+\sin(2y-z-x)+\sin(2z-x-y)=0$ and splitting the argument into $x-y, y-z, z-x$ terms, we get:
$$ \big[\sin(x-y)\cos(z-x)-\cos(x-y)\sin(z-x)\big] + \\ \big[\sin(y-z)\cos(x-y)-\cos(y-z)\sin(x-y)\big] + \\ \big[\sin(z-x)\cos(y-z)-\cos(z-x)\sin(y-z)\big] = 0$$
Grouping the terms according the cosine terms: $$ \cos(x-y)\big[\sin(y-z)-\sin(z-x)\big] + \\ \cos(y-z)\big[\sin(z-x)-\sin(x-y)\big] + \\ \cos(z-x)\big[\sin(x-y)-\sin(y-z)\big] = 0 \\ \\ \Leftrightarrow \\ \cos(x-y)\big[\sin(y)\cos(z)-\sin(z)\cos(y)-\sin(z)\cos(x)+\sin(x)\cos(z)\big] + \\ \cos(y-z)\big[\sin(z)\cos(x)-\sin(x)\cos(z)-\sin(x)\cos(y)+\sin(y)\cos(x)\big] + \\ \cos(z-x)\big[\sin(x)\cos(y)-\sin(y)\cos(x)-\sin(y)\cos(z)+\sin(z)\cos(y)\big] = 0 \\ \Leftrightarrow \\ \cos(x-y)\big[\big(\sin(x)+\sin(y)\big)\cos(z)-\big(\cos(x)+\cos(y)\big)\sin(z)\big] + \\ \cos(y-z)\big[\big(\sin(y)+\sin(z)\big)\cos(x)-\big(\cos(y)+\cos(z)\big)\sin(x)\big] + \\ \cos(z-x)\big[\big(\sin(z)+\sin(x)\big)\cos(y)-\big(\cos(z)+\cos(x)\big)\sin(y)\big] = 0 \\ $$
Taking one term from the rectangled parenthesis we se the following:
$$\big(\sin(x)+\sin(y)\big)\cos(z)-\big(\cos(x)+\cos(y)\big)\sin(z) = \\ \big(-\sin(z)\big)\cos(z)-\big(-\cos(z)\big)\sin(z) = -\sin(z)\cos(z)+\cos(z)\sin(z) = 0$$
It is similar for the other terms. So we get also that all terms are zero and therefore the sum also equals to zero. Q.E.D.
