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Let $x,y,z$ be real numbers such that $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$.

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  • $\begingroup$ Very nice exercise! $\endgroup$ – Peter Jan 9 '14 at 15:05
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Set

$$u:=cosx , v:=cosy , w:=cosz$$

We have

$$u + v + w = 0$$

and

$$cos3x + cos3y + cos3z = 4u^3 - 3u + 4v^3 - 3v + 4w^3 - 3w = 4u^3 + 4v^3 + 4w^3 = 0$$

So

$$u^3 + v^3 + w^3 = 0$$

Now consider

$$cos2x * cos2y * cos2z = (2u^2-1)(2v^2-1)(2w^2-1)$$

Further, we have

$$(u+v+w)^3 - u^3 - v^3 - w^3 = 3(u^2v+u^2w+v^2u+v^2w+w^2u+w^2v+2uvw) = 0$$

$$u^2v+u^2w+u^3+v^2u+v^2w+v^3+w^2u+w^2v+w^3+2uvw = 0$$

$$2uvw = 0$$

WLOG u = 0 , v=-w

As $2v^2-1=2w^2-1$ and $2u^2-1 < 0$, the proof is completed.

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Setting $\cos x=a$ etc.

We have $\displaystyle a+b+c=0\ \ \ \ (1)$ $\displaystyle\implies a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3+-3ab(-c)+c^3$ $\displaystyle\implies a^3+b^3+c^3=3abc\ \ \ \ (2)$

Again as $\displaystyle\cos3x=4\cos^3x-3\cos x\implies,$

$\sum\cos3x=0\implies 4(a^3+b^3+c^3)=3(a+b+c)=0\ \ \ \ (3)$

From $\displaystyle (2),(3) 3abc=0$

$\displaystyle\implies $ at least one of $a,b,c$ is zero

If $c=0,$ from $(1), a+b=0\iff b=-a$

$\displaystyle\implies\cos2x\cos2y\cos2z=\prod(2\cos^2x-1)=-(2\cos^2x-1)^2\le0$ as $\cos x=-\cos y\implies \cos^2x=\cos^2y$

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