×

# 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

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. 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 1paragraph 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} $$\boxed{123}$$

Sort by:

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

- 3 years, 6 months ago

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.

- 3 years, 6 months ago

One can also prove that-

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

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

- 3 years, 6 months 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...

- 3 years, 6 months ago

it can be prooved by usng complex numbers

- 3 years, 6 months ago

why y=a-120

- 3 years, 6 months ago

- 3 years, 6 months ago

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

Staff - 3 years, 6 months ago