Deducir la Matriz de Rotación 2D y Encontrar Ayuda

Zipi Damn
Messages
11
Reaction score
0
I was trying to deduce the 2D Rotation Matrix and I got frustrated. So, I found this article: Ampliación del Sólido Rígido/ (in Spanish).

rotacic3b3n-vectores.jpg



I don't understand the second line. How does he separate the matrix in two different parts?
Thanks for your time.
 
Physics news on Phys.org
Hi

In the matrix product in the second line, the vector (cos(sigma + phi), sin(sigma+phi)) should be (cos(sigma), sin(sigma)), which when multiplied by R is by definition (x,y).

Hope this helps.
 
  • Like
Likes 1 person
I don't know why he uses cos(σ+ψ) and sin(σ+ψ) instead of cos(σ) and sin(σ) when the matrix of the second line is separated.

That would make cos(σ+ψ)=cos(σ). Is this true? I can't see that relation. Because there is no similarity between the triangles formed by the vector (x,y) and the vector (x',y'). So it's imposible the cosine is the same.
 
I think it's just a mistake to be honest. It's definitely not true that cos(sigma + phi)=cos(sigma) for all values of these variables, so I think it's safe to assume it's just a mistake.
 
traxter said:
I think it's just a mistake to be honest. It's definitely not true that cos(sigma + phi)=cos(sigma) for all values of these variables, so I think it's safe to assume it's just a mistake.

Yes, it seems to be a mistake. But this mistake has helped me to analize better these concepts.
Anyway, thank you!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top