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I am half stumped on this rather confusing problem: 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 $\cos 2x + \cos 2y + \cos 2z = \sin 2x + \sin 2y + \sin 2z = 0$.

Follow the hint below, I figured out that a + b+ c = 0.

Let $a = e^{ix}$, $b = e^{iy}$, and $c = e^{iz}$. What do the given equations tell you about a, b, and c? How can you relate a, b, and c to what you want to prove?

Now how do I proceed? I tried using the double angle formula but it doesn't work :(Help is appreciated, thank you.

Freedom
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    What are the real and imaginary parts of $a^2, b^2,$ and $c^2$? – John Hughes Jan 19 '14 at 04:01
  • cos 2x and sin 2x right? So are you trying to tell me to use De moivre's theorem? thanks – Freedom Jan 19 '14 at 04:11
  • Well, you've got information about $a, b,$ and $c$, and you need to prove something about $a^2, b^2,$ and $c^2$. If, for instance, you could prove that $a^2 + b^2 + c^2 = 0$, you'd be done. I have half of an idea for proving this, but not all the details. But it seems likely that working in the complex plane is easier than in the real line for this problem. A suggestion: turn $a + b + c = 0$ into $a^{-1}a + a^{-1}b + a^{-1}c = 0$, i.e., $1 + d + e = 0$ for some complex numbers $d$ and $e$. That means that $d$ and $e$ must be complex conjugates. See where that takes you. – John Hughes Jan 19 '14 at 04:40

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If $\displaystyle a=\cos x+i\sin x$ etc., we have $\displaystyle a+b+c=0\ \ \ \ (1)$

and $\displaystyle \frac1a=\frac1{\cos x+i\sin x}=\cos x-i\sin x$

$\implies\displaystyle\frac1a+\frac1b+\frac1c=0\iff ab+bc+ca=0\ \ \ \ (2)$

Use $(1),(2)$ in the following identity: $\displaystyle (a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)$

lab bhattacharjee
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