If

\(cos \alpha + cos \beta +cos \gamma = sin \alpha + sin \beta +sin \gamma = 0 \)

Then Prove that

\(cos2 \alpha + cos2 \beta +cos2 \gamma = sin2 \alpha + sin2 \beta +sin2 \gamma = 0 \)

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## Comments

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TopNewestHi there! Besides complex numbers and/or vectors, I guess a purely trigo proof would be interesting as well (don't worry, it actually isn't at all that messy either...)

So we are given that \( \sin \gamma = - \sin \alpha - \sin \beta \) and \( \cos \gamma = - \cos \alpha - \cos \beta \). Then, using the identity \( \sin^2 \gamma + \cos^2 \gamma = 1\), and expanding, this results in \( 2 + 2 \sin \alpha \sin \beta + 2 \cos \alpha \cos \beta = 1 \), and simplifying, we may obtain that \( \cos ( \alpha - \beta ) = - \frac{1}{2} \), or in other words, the difference between \( \alpha \) and \( \beta \) is 120 degrees. Similarly, since the angles are symmetric, we can say that the difference between \( \beta \) and \( \gamma \), as well as the difference between \( \gamma \) and \( \alpha \), are 120 degrees as well. In other words, when drawn as the angle from the positive x-axis on cartesian axes, these represent 3 equally spaced angles through the 'cycle' of 360 degrees.

Now, we may then WLOG assume \( \beta = \alpha + 120 ^ \circ \) and \( \gamma = \alpha - 120 ^ \circ \).

Hence, \( \sin 2 \alpha + \sin 2 \beta + \sin 2 \gamma \) \( = \sin 2 \alpha + \sin ( 2 \alpha + 240 ^ \circ ) + \sin ( 2 \alpha - 240 ^ \circ ) \) \( = \sin 2 \alpha + 2 \sin 2 \alpha \cos 240 ^ \circ \) \( = \sin 2 \alpha - \sin 2 \alpha = 0 \)

Similarly, \( \cos 2 \alpha + \cos 2 \beta + \cos 2 \gamma \) \( = \cos 2 \alpha + \cos ( 2 \alpha + 240 ^ \circ ) + \cos ( 2 \alpha - 240 ^ \circ ) \) \( = \cos 2 \alpha + 2 \cos 2 \alpha \cos 240 ^ \circ \) \( = \cos 2 \alpha - \cos 2 \alpha = 0 \)

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This problem has a familiar ring to it, doesn't it? Something about complex numbers (or maybe vectors). But I don't want to spoil it with the solution.

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One can also prove that-

\( cos3α + cos3β + cos3ϒ = 3 ~cos(α + β + ϒ) \)

\( sin3α + sin3β + sin3ϒ = 3 ~sin(α + β + ϒ) \)

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Right, we want to show that if \( |a|=|b|=|c| = 1 \) and \( a+b+c = 0 \) then \( a^2+b^2+c^2 = 0 \). Pretty straightforward. There is also a nice connection with equilateral triangles...

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it can be prooved by usng complex numbers

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why y=a-120

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How do you answer this?

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Ah very nice. I set this question in the practice section. It's an interesting result.

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