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

Definition

For rotation by an angle $\theta$ about the origin:

• $M:\eq\left(\begin{array}{cc}\cos(\theta)&-\sin(\theta)\\ \sin(\theta)& \cos(\theta)\end{array}\right)$

This can be seen rather obviously by noting that:

• $(1,0)^\mathrm{T} \mapsto (\cos(\theta),\sin(\theta))^\mathrm{T}$, and
• $(0,1)^\mathrm{T} \mapsto (-\sin(\theta),\cos(\theta))^\mathrm{T}$

Then invoking some theorem about a linear map is defined by it's actions on a basis

• TODO: Which theorem?

Page notes

Inverse

For $c:\eq\cos(\theta)$ and $s:\eq\sin(\theta)$ we see that:

• $$M^{-1}\eq \frac{1}{cc+ss}\left(\begin{array}{cc}c&s\\-s&c\end{array}\right)$$ by inverse of a 2x2 matrix
  • Then note that $cc+ss\eq c^2+s^2$ and recall that cos^2+sin^2=1

So it degenerates into:

• $$M^{-1}\eq \left(\begin{array}{cc}c&s\\-s&c\end{array}\right)$$
  • Notice that $\cos(-\theta)\eq\cos(\theta)$ and $\sin(-\theta)\eq-\sin(\theta)$ - so this is just the rotation matrix for $-\theta$!

Double angle formula

Let $c:\eq\cos(\theta)$, $s:\eq\sin(\theta)$, $c':\eq\cos(\phi)$ and $s':\eq\sin(\phi)$ and let $M_\theta$ denote the rotation matrix for a rotation by $\theta$ We notice that $M_\theta\cdot M_\phi$

• $\eq M_{\theta+\phi}$ by "addition of angles"

and this matrix must be equal to $M_\theta\cdot M_\phi$

By equating the entries we get the double angle formulas