×

# 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, 1 month ago

## 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, $$\cos 2 \alpha + \cos 2 \beta + \cos 2 \gamma$$ $$= \cos 2 \alpha + \cos ( 2 \alpha + 240 ^ \circ ) + \cos ( 2 \alpha - 240 ^ \circ )$$ $$= \cos 2 \alpha + 2 \cos 2 \alpha \cos 240 ^ \circ$$ $$= \cos 2 \alpha - \cos 2 \alpha = 0$$ · 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. · 3 years, 1 month ago

Log in to reply

One can also prove that-

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

$$sin3α + sin3β + sin3ϒ = 3 ~sin(α + β + ϒ)$$ · 3 years, 1 month 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... · 3 years, 1 month ago

Log in to reply

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

Log in to reply

why y=a-120 · 3 years, 1 month ago

Log in to reply

How do you answer this? · 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. Staff · 3 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...