Trigonometric equation with complex numbers

Question

Let $x$, $y$, and $z$ be real numbers such that $\cos x+\cos y+\cos z=\sin x+\sin y+\sin z=0$.

Prove that $\cos 2x+\cos 2y+\cos 2z=\sin 2x+\sin 2y+\sin 2z=0$.

Starting with the given equation, I got that $i\sin x+i\sin y+i\sin z=0$.

Adding this to the other part of the given equation, I then got $\cos x+\cos y+\cos z+i\sin x+i\sin y+i\sin z=0$, which can also be written as $(\cos x+i\sin x)+(\cos y+i\sin y)+(\cos z+i\sin z)=0$.

Here, I let $a=e^{ix}$, $b=e^{iy}$, and $c=e^{iz}$, which, after substituting in to the above equation gives $a+b+c=0$.

What we want to prove can be written as $a^2+b^2+c^2$, but I am not quite sure how to find this result from what I have.

Some help would be very appreciated. Thanks!

Answer 1

Multiply the equations by $\sin(x+y+z)$ and $\cos(x+y+z)$ and subtract them to get

$$\sin(x+y+z)\cos x -\cos(x+y+z)\sin x +\\ \sin(x+y+z)\cos y -\cos(x+y+z)\sin y +\\ \sin(x+y+z)\cos z -\cos(x+y+z)\sin z\\ =\sin(y+z)+\sin(x+z)+\sin(x+y)=0$$ In a similar way we get $$\cos(y+z)+\cos(x+z)+\cos(x+y)=0$$

Now if we square both equations and subtract them we have $$\cos 2x+ \cos 2y+ \cos 2z +2\cos(y+z)+2\cos(x+z)+2\cos(x+y)=0$$ and if we multiply the two equations together we have

$$\sin 2x+ \sin 2y+ \sin 2z +2\sin(y+z)+2\sin(x+z)+2\sin(x+y)=0$$

Answer 2

In the same way you have: $(\cos x-i \sin x)+(\cos y-i\sin y)+(\cos z-i \sin z)=0$, it's equal: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, so $ab+ac+bc=0$ and $0=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2=0$