If  $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$ Where  $\theta =\frac{2\pi }{19}$ then  $A^{2017}=$

Given that  $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\theta =\frac{2\pi }{19}$

$\therefore A^n=\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$

$A^{2017}=\left[\begin{array}{cc}\cos 2017\theta & \sin 2017\theta \\ -\sin 2017\theta & \cos 2017\theta \end{array}\right]$

$2017\theta =2017\text{x}\frac{2\pi }{19}=\left(106+\frac{3}{19}\right)2\pi$

$A^{2017}=\left[\begin{array}{cc}\cos 3\theta & \sin 3\theta \\ -\sin 3\theta & \cos 3\theta \end{array}\right]=A^3$