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$\cos(A-B) + \cos(B-C) + \cos(C-A) = \frac{-3}{2}$ We need to prove that $\cos A + \cos B + \cos C = \sin A + \sin B + \sin C = 0$

I was wondering if it's possible to prove this result by showing that the real and imaginary parts of $z = \cos A + \cos B + \cos C$ are equal to zero, somehow invoking Vieta's or De Moivre's theorem if required. I tried, starting with $\cos(B-C)$ and other cyclic terms but couldn't really get anywhere.

Any other method is also appreciated.

Thanks a lot!

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Let $u$, $v$, $w$ be $e^{iA}$, $e^{iB}$ and $e^{iC}$. Then $$(u+v+w)(u^{-1}+v^{-1}+w^{-1}) =2\cos(A-B)+2\cos(B-C)+2\cos(C-A)+3.$$ Your condition is equivalent to $(u+v+w)(u^{-1}+v^{-1}+w^{-1})=0$. These brackets are complex conjugates, so $u+v+w=0$ and this means $$(\cos A+\cos B+\cos C)+i(\sin A+\sin B+\sin C)=0.$$

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  • $\begingroup$ Great man, thanks a lot. I liked that! Just a question, are there any other trigonometry problems you know of, which can be solved using complex numbers? $\endgroup$ – arya_stark Jul 1 '17 at 15:45
  • $\begingroup$ a good and short proof! $\endgroup$ – Dr. Sonnhard Graubner Jul 1 '17 at 15:51
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    $\begingroup$ @user28968: Napoleon's theorem, for instance - en.wikipedia.org/wiki/Napoleon%27s_theorem $\endgroup$ – Jack D'Aurizio Jul 1 '17 at 15:56

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