Rotation Transformation

**bugatti79** · October 16th 2012, 11:21 AM

Folks,

I am struggling to understand how the attached expressions were obtained for the simple rotation of 2 axes normal to each other by an angle $\theta$

This is different to what is shown the wiki link Rotation matrix - Wikipedia, the free encyclopedia (See in 'two dimensions')

Ie, $x'= x\cos \theta - y \sin \theta$ and $y'=x \sin \theta + y \cos \theta$

Any clues?

Regards
bugatti

**TheEmptySet** · October 16th 2012, 12:00 PM

Originally Posted by bugatti79

Folks,

I am struggling to understand how the attached expressions were obtained for the simple rotation of 2 axes normal to each other by an angle $\theta$

This is different to what is shown the wiki link Rotation matrix - Wikipedia, the free encyclopedia (See in 'two dimensions')

Ie, $x'= x\cos \theta - y \sin \theta$ and $y'=x \sin \theta + y \cos \theta$

Any clues?

Regards
bugatti

Here is a picture

Note that

$T(e_1)=\cos(\theta)e_1+\sin(\theta)e_1$

and

$T(e_2)=-\sin(\theta)e_1+\cos(\theta)e_2$

So the general transformation is

$A=\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$

Now for any arbitary x and y you get

$A \begin{bmatrix} x \\ y \end{bmatrix} =\begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} x \cos(\theta)-y \sin(\theta) \\ x \sin(\theta)+y \cos(\theta) \end{bmatrix}$

**bugatti79** · October 16th 2012, 12:16 PM

Thank you sir, just what I was looking for.

Regards

**johnsomeone** · October 16th 2012, 12:18 PM

I wrote this out for someone last month, if you want to take a look:

Understanding how to express basis with trig functions

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