If

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

Then Prove that

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

If

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

Then Prove that

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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, \( \cos 2 \alpha + \cos 2 \beta + \cos 2 \gamma \) \( = \cos 2 \alpha + \cos ( 2 \alpha + 240 ^ \circ ) + \cos ( 2 \alpha - 240 ^ \circ ) \) \( = \cos 2 \alpha + 2 \cos 2 \alpha \cos 240 ^ \circ \) \( = \cos 2 \alpha - \cos 2 \alpha = 0 \) – Jau Tung Chan · 3 years, 1 month ago

Log in to reply

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. – Michael Mendrin · 3 years, 1 month ago

Log in to reply

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

\( sin3α + sin3β + sin3ϒ = 3 ~sin(α + β + ϒ) \) – Avineil Jain · 3 years, 1 month ago

Log in to reply

– Patrick Corn · 3 years, 1 month ago

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...Log in to reply

it can be prooved by usng complex numbers – Dharma Teja · 3 years, 1 month ago

Log in to reply

why y=a-120 – Yee Cheng · 3 years, 1 month ago

Log in to reply

How do you answer this? – Kent Mercado · 3 years, 1 month ago

Log in to reply

Ah very nice. I set this question in the practice section. It's an interesting result. – Calvin Lin Staff · 3 years, 1 month ago

Log in to reply