2-Dimensional rotation matrix
Core concepts:
- Definition
- Inverse is again a rotation
- Alternate proof of double angle formulas
Definition
For rotation by an angle [ilmath]\theta[/ilmath] about the origin:
- [ilmath]M:\eq\left(\begin{array}{cc}\cos(\theta)&-\sin(\theta)\\ \sin(\theta)& \cos(\theta)\end{array}\right)[/ilmath]
This can be seen rather obviously by noting that:
- [ilmath](1,0)^\mathrm{T} \mapsto (\cos(\theta),\sin(\theta))^\mathrm{T} [/ilmath], and
- [ilmath](0,1)^\mathrm{T} \mapsto (-\sin(\theta),\cos(\theta))^\mathrm{T} [/ilmath]
Then invoking some theorem about a linear map is defined by it's actions on a basis
- TODO: Which theorem?
Page notes
Inverse
For [ilmath]c:\eq\cos(\theta)[/ilmath] and [ilmath]s:\eq\sin(\theta)[/ilmath] we see that:
- [math]M^{-1}\eq \frac{1}{cc+ss}\left(\begin{array}{cc}c&s\\-s&c\end{array}\right)[/math] by inverse of a 2x2 matrix
- Then note that [ilmath]cc+ss\eq c^2+s^2[/ilmath] and recall that cos^2+sin^2=1
So it degenerates into:
- [math]M^{-1}\eq \left(\begin{array}{cc}c&s\\-s&c\end{array}\right)[/math]
- Notice that [ilmath]\cos(-\theta)\eq\cos(\theta)[/ilmath] and [ilmath]\sin(-\theta)\eq-\sin(\theta)[/ilmath] - so this is just the rotation matrix for [ilmath]-\theta[/ilmath]!
Double angle formula
Let [ilmath]c:\eq\cos(\theta)[/ilmath], [ilmath]s:\eq\sin(\theta)[/ilmath], [ilmath]c':\eq\cos(\phi)[/ilmath] and [ilmath]s':\eq\sin(\phi)[/ilmath] and let [ilmath]M_\theta[/ilmath] denote the rotation matrix for a rotation by [ilmath]\theta[/ilmath] We notice that [ilmath]M_\theta\cdot M_\phi[/ilmath]
- [ilmath]\eq M_{\theta+\phi} [/ilmath] by "addition of angles"
and this matrix must be equal to [ilmath]M_\theta\cdot M_\phi[/ilmath]
By equating the entries we get the double angle formulas
