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

$A=\begin{bmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{bmatrix}$. Let $P(n): A^n=\begin{bmatrix}\cos n\theta & -\sin n\theta\\\sin n\theta & \cos n\theta\end{bmatrix}$.

Base case ($n=1$): RHS $=\begin{bmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{bmatrix}=A$. $P(1)$ holds.

Inductive step: Assume $P(k)$: $A^k=\begin{bmatrix}\cos k\theta & -\sin k\theta\\\sin k\theta & \cos k\theta\end{bmatrix}$.

$A^{k+1}=A^k\cdot A$: entry(1,1) $=\cos k\theta\cos\theta-\sin k\theta\sin\theta=\cos((k+1)\theta)$.

entry(1,2) $=\cos k\theta(-\sin\theta)+(-\sin k\theta)\cos\theta=-[\sin k\theta\cos\theta+\cos k\theta\sin\theta]=-\sin((k+1)\theta)$.

entry(2,1) $=\sin k\theta\cos\theta+\cos k\theta\sin\theta=\sin((k+1)\theta)$. …

Prove by mathematical induction using the angle-addition for …

• Sign errors applying the sine/cosine angle-addition formulas. …