Trigo-no-metry

$cos \alpha + cos \beta +cos \gamma = sin \alpha + sin \beta +sin \gamma = 0 $

Then Prove that

$cos2 \alpha + cos2 \beta +cos2 \gamma = sin2 \alpha + sin2 \beta +sin2 \gamma = 0$

#Algebra #Trigonometry #Proofs

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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

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 - 7 years 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.

Michael Mendrin - 7 years ago

One can also prove that-

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

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

Avineil Jain - 7 years 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...

Patrick Corn - 7 years ago

it can be prooved by usng complex numbers

Dharma Teja - 7 years ago

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

Calvin Lin Staff - 7 years ago

How do you answer this?

Kent Mercado - 7 years ago

why y=a-120

yee cheng - 7 years ago

Math	Appears 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}`	$\boxed{123}$