# Insight/Intuition into rotations in R²

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1. Feb 24, 2016

### STENDEC

I've been using rotation matrices for quite some time now without fully grasping them. Whenever I tried to develop an intuitive understanding of...$$x' = x\cos\theta - y\sin\theta \\ y' = x\sin\theta + y \cos\theta$$... I failed and gave up. I've looked at numerous online texts and videos, but following the step-by-step explanations didn't lead to me seeing the whole picture as I had hoped.

Could someone explain to me (like I'm 5 years old), why $-y\sin\theta$ and $x\sin\theta$ are used to affect the value along the other axis?

Looking at the following picture (pardon the quality):

Is the contribution of $y$ to $x$ and vice versa there, to ensure that $P$ maintains the correct distance to the origin, or is that a misguided simplification of mine? The yellow line cannot be $sin + cos$ (Pythagorean theorem) yet I may combine these two to get $x'$ and $y'$. Do you see where my gap in understanding lies? Is there a drawing that could clarify how these terms combine to give the correct value we observe? Algebraic proofs don't work with me I'm afraid, I need a geometric/visual explanation.

Last edited: Feb 24, 2016
2. Feb 24, 2016

### Staff: Mentor

MAYBE THIS VIDEO WILL HELP:

3. Feb 25, 2016

### Edgardo

4. Feb 28, 2016

### suremarc

The component $\left(\begin{smallmatrix}x\cos\theta\\y\cos\theta\end{smallmatrix}\right)$ represents the projection of $\left(\begin{smallmatrix}x'\\y'\end{smallmatrix}\right)$ onto the original vector $\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)$. The other component (the "rejection") should therefore have magnitude $|\sqrt{x^2+y^2}\sin\theta|$ and be orthogonal to the projection. There are two unique solutions for such a vector.

Finally, when $\theta$ is nonzero but less than a straight angle, an increase in $y'$ corresponds to a decrease in $x'$; i.e., the x-component of the rejection should be negative. (This condition is equivalent to a choice of orientation on $\mathbb{R}^2$.) This tells us which solution is the desired one.

As for your drawing, I don't quite understand what you mean. Could you elaborate?