15-00-DerivationOfRotationMatrixUsingPolarCoordinates.tex

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\pmcreated{2013-03-22 15:25:02}
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\pmtitle{derivation of rotation matrix using polar coordinates}
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\pmtype{Derivation}
\pmcomment{trigger rebuild}
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%\pmkeywords{rotation matrix}
\pmrelated{RotationMatrix}
\pmrelated{PolarCoordinates}
\pmrelated{DecompositionOfOrthogonalOperatorsAsRotationsAndReflections}
\pmrelated{DerivationOf2DReflectionMatrix}

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\begin{document}
We derive formally the expression for the rotation of a two-dimensional vector
$\vv = a\vx + b\vy$ by an angle $\phi$ counter-clockwise.  Here 
$\vx$ and $\vy$ are perpendicular unit vectors that are oriented counter-clockwise
(the usual orientation).

\begin{center}
\includegraphics{rotation-matrix.eps}
\end{center}

In terms of polar coordinates, $\vv$ may be rewritten:
\begin{align*}
\vv &= r(\cos \theta \, \vx + \sin \theta \, \vy) \,, \quad a = r\cos \theta\,; b = r\sin\theta\,, \\
\intertext{for some angle $\theta$ and radius $r \geq 0$.
To rotate a vector $\vv$ by $\phi$ really means to shift its
polar angle by a constant amount $\phi$ but leave its polar radius fixed.
Therefore, the result of the rotation must be:}
\vv' &= r\bigl(\cos(\theta + \phi) \, \vx + \sin(\theta + \phi) \, \vy\bigr) \\
\intertext{Expanding using the angle addition formulae, we obtain}
\vv' &= r\bigl(\cos \theta \cos \phi - \sin \theta \sin \phi) \, \vx + (\sin \theta \cos \phi + \cos \theta \sin \phi) \, \vy\bigr)  \\
&= (a \cos \phi - b \sin \phi) \, \vx + (b \cos \phi + a \sin \phi) \, \vy\,.
\end{align*}
When this transformation is written out in $[\vx, \vy]$-coordinates, we obtain the formula for the rotation matrix:
\[
\vv' = \begin{bmatrix}
\cos \phi & -\sin \phi \\
\sin \phi & \cos \phi
\end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} \,.
\]
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\end{document}