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This question today appeared in my maths olympiad paper:

If $\cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0$, then, prove that $\cos 2x + \cos 2y + \cos 2z = \sin 2x + \sin 2y + \sin 2z = 0$.

Can anyone please help me in finding out the solution of this equation ?

I have not gone any far in this solution.

Daniel R
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Vishwesh
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1 Answers1

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Putting $a=\cos x+i\sin x$ etc,

we have $a+b+c=0$

and $a^{-1}=\frac1{\cos x+i\sin x}=\cos x-i\sin x$ $\implies a^{-1}+b^{-1}+c^{-1}=0\implies ab+bc+ca=0$

$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0$

Now, $a^2=(\cos x+i\sin x)^2=\cos^2x-\sin^2x+i2\sin x\cos x=\cos2x+i\sin2x$ which is a special case of de Moivre's formula

lab bhattacharjee
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  • Brilliant answer ! Just one question, is there any other trigonometric proof or will this proof be accepted in IMO or any other such mathematical olympiad. – Vishwesh Sep 01 '13 at 12:57
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    @Vishwesh, have you understood the method? I think any logical method should be accepted in such competition – lab bhattacharjee Sep 01 '13 at 13:02
  • Yes @lab bhattacharjee, I had the idea of de Moivre's theorem but could never think of applying it here. Fantastic method and flawless work. – Vishwesh Sep 01 '13 at 13:05
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    @Vishwesh, a related problem : http://math.stackexchange.com/questions/479726/if-cos-x-2-cos-y3-cos-z-0-sin-x2-sin-y3-sin-z-0-and-xyz-pi – lab bhattacharjee Sep 01 '13 at 13:07
  • Ok, so if, nothing works, polar form can mostly come for rescue provided that solution must involve cos or sin. – Vishwesh Sep 01 '13 at 13:09