2-Dimensional rotation matrix

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Demote once core concepts are covered. Alec (talk) 08:07, 3 April 2018 (UTC)

Core concepts:

  1. Definition
  2. Inverse is again a rotation
  3. Alternate proof of double angle formulas


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


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