Math Book Cambodia: It is given three positive real number `x,y,z` such that : `cos x+cosy+cosz=0` and `cos3x+cos3y+cos3z=0`

Friday, November 27, 2020

It is given three positive real number $x,y,z$ such that : $cos x+cosy+cosz=0$ and $cos3x+cos3y+cos3z=0$ .

Prove that: $cos 2x.cos 2y.cos 2z$ ≤ 0

Solution

Prove that : $cos 2x.cos 2y.cos 2z$ ≤ 0

Following formula :

$\cos3a=4\cos^3a-3\cos a$ then we can see that:

$\cos3x=4\cos^3x-3\cos x$ (1)

$\cos3y=4\cos^3y-3\cos y$ (2)

$\cos3z=4\cos^3z-3\cos z$ (3)

Make the sum of 3 equations above side to side, we will get:
$4\cos^3x+4\cos^3y+4\cos^3z=0$ because $cos x+cosy+cosz=0$ and $cos3x+cos3y+cos3z=0$

From $cos x+cosy+cosz=0$ then $cos x+cosy=-cosz$

Let's make the 3rd power to the equation above:
$(cosx+cosy)^3=-cos^3z$

$cos^3z+3cosxcosy(cosx+cosy)+cos^3y=-cos^3z$

$cos^3z-3cosxcosycosz+cos^3y=-cos^3z$

$cos^3z+cos^3y+cos^3z =3cosxcosycosz$

But, we knew that: $cos3x+cos3y+cos3z=0$

Then, $3cosxcosycosz=0$ ⇒ $cosx=0$ or $cosy=0$ or $cosz=0$

We assume that: $cosx=0$ and $cosy=-cosz$

Then, $cos2xcos2ycos2z=(2cos^2x-1)(2cos^2y-1)(2cos^2z-1)$

$cos2xcos2ycos2z=-(2cos^2z-1)^2$ ≤ 0

Hence, the problem is proved.

Solution by: Thin Sokkean

Math Book Cambodia