We shall prove the result by using principle of mathematical induction.

We have   $\mathrm{P}(n):$ If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right],$ then $\mathrm{A}^{n}=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right], n \in {N}$

$P(1): A=\left[\begin{array}{ll}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right],$ so $A^{1}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]$

Therefore, the result is true for $n=1$

Let the result be true for $n=k$. So

$\mathrm{P}(k): \mathrm{A}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right],$ then $\mathrm{A}^{k}=\left[\begin{array}{cc}\cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta\end{array}\right]$

Now, we prove that the result holds for $n=k+1$

Now                     $A^{k+1}=A \cdot A^{k}=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]\left[\begin{array}{cc}\cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta\end{array}\right]$

$=\left[\begin{array}{cc}\cos \theta \cos k \theta-\sin \theta \sin k \theta & \cos \theta \sin k \theta+\sin \theta \cos k \theta \\ -\sin \theta \cos k \theta+\cos \theta \sin k \theta & -\sin \theta \sin k \theta+\cos \theta \cos k \theta\end{array}\right]$

$=\left[\begin{array}{cc}\cos (\theta+k \theta) & \sin (\theta+k \theta) \\ -\sin (\theta+k \theta) & \cos (\theta+k \theta)\end{array}\right]=\left[\begin{array}{cc}\cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta\end{array}\right]$

Therefore, the result is true for $n=k+1 .$ Thus by principle of mathematical induction,

we have $A^{n}=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right],$ holds for all natural numbers.