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 $x,y,z \in [-\pi,\pi]$

Question

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?

$z=-(x+y)$ therefore $\frac{1-\cos x}{\sin x}=-\frac{1-\cos y}{\sin y}$ — Behrouz Maleki, Jan 04 '17 at 12:27
$$\sin x+\sin y +\sin(-(x+y))=\sin x+\sin y-\sin(x+y)=0$$ and $$\cos x+\cos y +\cos(-(x+y))=\cos x+\cos y+\cos(x+y)=0$$ — Behrouz Maleki, Jan 04 '17 at 12:33
Related : http://math.stackexchange.com/questions/1397066/clarification-regarding-a-question — lab bhattacharjee, Jan 04 '17 at 12:36
@labbhattacharjee I want a high school solution I mean without using complex numbers. — Taha Akbari, Jan 04 '17 at 12:41

score 1 · Answer 1 · answered Jan 04 '17 at 13:26

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?

Nosrati · Accepted Answer · 2017-01-05T09:38:18.393

1

\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 $$\frac{8\pi^2}{9}$$

edited Jan 05 '17 at 09:38

answered Jan 04 '17 at 13:41

Nosrati

29,995

So go ahead・・・. From here how do we go? – Takahiro Waki Jan 04 '17 at 13:59
@TakahiroWaki I can improve it. – Nosrati Jan 04 '17 at 14:50
Does their (x,y,z) satisfy OP's equation? The permutations of $(0,-2\pi/3,2\pi/3)$ has already 6 answers. – Takahiro Waki Jan 05 '17 at 00:02
@TakahiroWaki These equations are symmetric over triple $(x,y,z)$ by permutation and have $4\times3=12$ answers, but for this symmetrical property, the value of $x^2+y^2+z^2$ repeat three times. – Nosrati Jan 05 '17 at 04:36
Please show me specific (x,y,z). If so, I will upvote. – Takahiro Waki Jan 05 '17 at 05:07
Sorry because $cos-\pi/6+cos\pi/6+cos0=1+\sqrt3$・・・. – Takahiro Waki Jan 05 '17 at 09:14
@TakahiroWaki Thank you. I didn't check them truly. You are right. Indeed Only One case (4 case symmetrically) is accepted. I will correct it. – Nosrati Jan 05 '17 at 09:26

Takahiro Waki · Answer 3 · 2017-01-05T00:19:03.617

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, $$\displaystyle x^2+y^2+z^2=\frac{8\pi^2}9 $$

score 0 · Answer 4 · answered Jan 04 '17 at 15:59

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 $t\in [-\pi,\pi]$.

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 $x,y,z \in [-\pi,\pi]$

4 Answers4