If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$ THE PROBLEM: If $a\cos(θ)=b$ $\cos( θ+2π/3)=c\cos(θ+4π/3)$, prove that $ab+bc+ca=0.$
MY THOUGHT PROCESS: We have to prove that $ab+bc+ca=0$. 
One method using which we can do this is, if we can somehow obtain the equation $k(ab+bc+ca)=0$ we can deduce that $ab+bc+ca=0$ using the zero product rule. 
Another method to do this would be to obtain an expression which has $(ab+bc+ca)$ in it. The identity $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ has $2(ab+bc+ca)$ in it. Therefore if we can somehow show that $(a+b+c)^2=a^2+b^2+c^2$ our task would be over.
MY ATTEMPT: I have proved the result using the first approach. I tried to do it using the second one but could not proceed far. I was facing problems showing that $(a+b+c)^2=a^2+b^2+c^2$. If i get any help i shall be very grateful.
 A: I don't know whether I am just repeating your first solution.
Let $a\cos\theta=k$. Then
\begin{align*}
k(ab+bc+ca)&=abc\cos\left(\theta+\frac{4\pi}{3}\right)+abc\cos\theta+abc\cos\left(\theta+\frac{2\pi}{3}\right)\\
&=abc\left[\cos\left(\theta+\frac{2\pi}{3}\right)+2\cos\left(\theta+\frac{2\pi}{3}\right)\cos\frac{2\pi}{3}\right]\\&=0
\end{align*}
A: Using the addition formulas:
$$2\cos\left(\theta+\frac{2\pi}{3}\right)=-\cos\theta-\sqrt3\sin\theta$$
$$2\cos\left(\theta+\frac{4\pi}{3}\right)=-\cos\theta+\sqrt3\sin\theta$$
Then on the one hand
$$\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)+\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)+\cos\theta\cos\left(\theta+\frac{2\pi}{3}\right)=$$
$$\frac{\cos^2\theta-3\sin^2\theta}{4}+\frac{-\cos^2\theta+\sqrt3\sin\theta\cos\theta}{2}+\frac{-\cos^2\theta-\sqrt3\sin\theta\cos\theta}{2}=$$
$$\frac{\cos^2\theta-3\sin^2\theta}{4}-\cos^2\theta=\frac{\cos^2\theta-3\sin^2\theta-4\cos^2\theta}{4}-\cos^2\theta=\frac{-3\cos^2\theta-3\sin^2\theta}{4}=-\frac{3}{4}$$
On the other hand
$$\left(\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2+\left(\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2+\left(\cos\theta\cos\left(\theta+\frac{2\pi}{3}\right)\right)^2=$$
$$\left(\frac{\cos^2\theta-3\sin^2\theta}{4}\right)^2+\left(\frac{-\cos^2\theta+\sqrt3\sin\theta\cos\theta}{2}\right)^2+\left(\frac{-\cos^2\theta-\sqrt3\sin\theta\cos\theta}{2}\right)^2=$$
$$\frac{\cos^4\theta-6\sin^2\cos^2+9\sin^4}{16}+\frac{\cos^4\theta-2\sqrt3\sin\theta\cos^3\theta+3\sin^2\theta\cos^2\theta}{4}+\frac{\cos^4\theta+2\sqrt3\sin\theta\cos^3\theta+3\sin^2\theta\cos^2\theta}{4}=$$
$$\frac{\cos^4\theta-6\sin^2\cos^2+9\sin^4}{16}+\frac{2\cos^4\theta+6\sin^2\theta\cos^2\theta}{4}=$$
$$\frac{\cos^4\theta-6\sin^2\cos^2+9\sin^4+8\cos^4\theta+24\sin^2\theta\cos^2\theta}{16}=$$
$$\frac{9}{16}(\cos^4\theta+2\sin^2\theta\cos^2\theta+\sin^4\theta)=\frac{9}{16}\left(\cos^2\theta+\sin^2\theta\right)^2=\frac{9}{16}.$$
If we square the first equation and subtract it from the second we get $0$. 
Since
$$\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)(a+b+c)=$$
$$a\left(\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)+\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)+\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)\right)$$
and
$$\left(\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2(a^2+b^2+c^2)=$$
$$a^2\left(\left(\cos\left(\theta+\frac{2\pi}{3}\right)\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2+\left(\cos\theta\cos\left(\theta+\frac{4\pi}{3}\right)\right)^2+\left(\cos\theta\cos\left(\theta+\frac{2\pi}{3}\right)\right)^2\right)$$
we conclude
$$(a+b+c)^2-(a^2+b^2+c^2)=0$$
and $$ab+bc+ca=0$$
A: In the figure below, taking $0 \leq \theta \leq 30^\circ$, we have
$$|\overline{OX}| = |\overline{OA}|\;\cos\theta \;=\; -|\overline{OB}|\;\cos(\theta + 120^\circ) \;=\; -|\overline{OC}|\;\cos(\theta + 240^\circ) = |\overline{OX^\prime}|$$
... where all lengths (and, later, angles and areas) are considered non-negative. To match the notation of the problem, we define
$$a := |\overline{OA}| = |\overline{OA^\prime}| \qquad b := -|\overline{OB}| \qquad c := -|\overline{OC}| \tag{1}$$

Reflecting $A$ in $O$ to get $A^\prime$, we have ...
$$|\triangle BOC| = |\triangle A^\prime OB| + |\triangle A^\prime OC| \tag{2}$$
Therefore ...
$$\frac12 |\overline{OB}| |\overline{OC}| \sin \angle BOC = \frac12 |\overline{OA^\prime}||\overline{OB}|\sin \angle A^\prime OB + \frac12 |\overline{OA^\prime}||OC|\sin \angle A^\prime OC \tag{3}$$
Clearing fractions, and re-writing with $a$, $b$, $c$ via $(1)$ ...
$$(-b)(-c) \sin 120^\circ = a(-b)\sin 60^\circ + a(-c)\sin 60^\circ \tag{4}$$
Since $\sin 60^\circ = \sin 120^\circ$, we can divide-through. Rearranging ...
$$a b + b c + c a = 0 \quad\square \tag{5}$$
