Waste less time on Facebook — follow Brilliant.
×

Trigo-no-metry

If

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

Then Prove that

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

Note by Sudipta Biswas
3 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Hi 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, 6 months 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, 6 months ago

Log in to reply

One can also prove that-

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

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

Avineil Jain - 3 years, 6 months ago

Log in to reply

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...

Patrick Corn - 3 years, 6 months ago

Log in to reply

it can be prooved by usng complex numbers

Dharma Teja - 3 years, 6 months ago

Log in to reply

why y=a-120

Yee Cheng - 3 years, 6 months ago

Log in to reply

How do you answer this?

Kent Mercado - 3 years, 6 months 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, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...