If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,then 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) We shall prove the result by using the principle of mathematical induction.
Let $P(n)$ be the statement: If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,then $A^{n} = \begin{bmatrix} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{bmatrix}$ for $n \in N$.
Step $1$: For $n = 1$,$A^{1} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$. This is true.
Step $2$: Assume the result is true for $n = k$. That is,$A^{k} = \begin{bmatrix} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{bmatrix}$.
Step $3$: We prove the result for $n = k + 1$.
$A^{k+1} = A \cdot A^{k} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta \end{bmatrix}$
$= \begin{bmatrix} \cos \theta \cos k \theta - \sin \theta \sin k \theta & \cos \theta \sin k \theta + \sin \theta \cos k \theta \\ -\sin \theta \cos k \theta - \cos \theta \sin k \theta & -\sin \theta \sin k \theta + \cos \theta \cos k \theta \end{bmatrix}$
Using trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$:
$A^{k+1} = \begin{bmatrix} \cos(k+1)\theta & \sin(k+1)\theta \\ -\sin(k+1)\theta & \cos(k+1)\theta \end{bmatrix}$.
Thus,the result holds for $n = k+1$. By the principle of mathematical induction,the statement is true for all $n \in N$.