Proof of the Power Formula for Rotation Transformation Matrices
Theorem
For every natural number $n$, the following holds. $$ \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix} ^{n} = \begin{bmatrix} { \cos n\theta }&{ -\sin n\theta } \\ { \sin n\theta }&{ \cos n\theta } \end{bmatrix} $$
Explanation
A linear transformation matrix that rotates by $\theta$ around the origin, when squared, results in a linear transformation that rotates by $n\theta$.
Proof
Strategy: It’s intuitively obvious, and can be easily proven using mathematical induction.
$$ (ㄱ) : \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix}^{ n }= \begin{bmatrix} { \cos n\theta }&{ -\sin n\theta } \\ { \sin n\theta }&{ \cos n\theta } \end{bmatrix} $$
When $n=1$, $$ \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix} = \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix} $$ Therefore, condition (ㄱ) is satisfied. Now, assuming condition (ㄱ) holds when $n=k$, $$ \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix}^{ k }= \begin{bmatrix} { \cos k\theta }&{ -\sin k\theta } \\ { \sin k\theta }&{ \cos k\theta } \end{bmatrix} $$ Multiplying both sides by $\begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta }\end{bmatrix}$ results in $$ \begin{align*} \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix}^{ k+1 } =& \begin{bmatrix} { \cos k\theta }&{ -\sin k\theta } \\ { \sin k\theta }&{ \cos k\theta } \end{bmatrix} \begin{bmatrix} { \cos \theta }&{ -\sin \theta } \\ { \sin \theta }&{ \cos \theta } \end{bmatrix} \\ =& \begin{bmatrix} { \cos k\theta \cos \theta -\sin k\theta \sin \theta }&{ -(\sin k\theta \cos \theta +\cos k\theta \sin \theta ) } \\ { \sin k\theta \cos \theta +\cos k\theta \sin \theta }&{ \cos k\theta \cos \theta -\sin k\theta \sin \theta } \end{bmatrix} \\ =& \begin{bmatrix} { \cos (k+1)\theta }&{ -\sin (k+1)\theta } \\ { \sin (k+1)\theta }&{ \cos (k+1)\theta } \end{bmatrix} \end{align*} $$ Therefore, condition (ㄱ) is satisfied.
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