Q.If $A = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, prove that $A^n = \begin{bmatrix} \cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{bmatrix}$ where $n \in N$.

Prove by mathematical induction on $n$. Base $n=1$ is immediate; the inductive step multiplies $A^k$ by $A$ and collapses entries with $\cos(k\theta\pm\theta)$, $\sin(k\theta\pm\theta)$ formulas.

Concept. Use the principle of mathematical induction, plus the angle-addition identities $\cos(k\theta+\theta)=\cos k\theta\cos\theta-\sin k\theta\sin\theta$ and $\sin(k\theta+\theta)=\sin k\theta\cos\theta+\cos k\theta\sin\theta$.

Base case $n=1$.

$A^1=\begin{bmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{bmatrix}=\begin{bmatrix}\cos 1\theta&\sin 1\theta\\ -\sin 1\theta&\cos 1\theta\end{bmatrix}. \checkmark$

Inductive hypothesis. Assume for $n=k$:

$A^k=\begin{bmatrix}\cos k\theta&\sin k\theta\\ -\sin k\theta&\cos k\theta\end{bmatrix}.$

Inductive step. Then

$A^{k+1}=A^k A=\begin{bmatrix}\cos k\theta&\sin k\theta\\ -\sin k\theta&\cos k\theta\end{bmatrix}\begin{bmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{bmatrix}.$

Compute entries:

• $(1,1)=\cos k\theta\cos\theta-\sin k\theta\sin\theta=\cos(k+1)\theta$,
• $(1,2)=\cos k\theta\sin\theta+\sin k\theta\cos\theta=\sin(k+1)\theta$, …