Question

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

• A. $\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$
• B. $\left[\begin{array}{cc}{\cos }^n\theta & {\sin }^n\theta \\ {(-1)}^n{\sin }^n\theta & {\cos }^n\theta \end{array}\right]$
• C. $\left[\begin{array}{cc}n\cos \theta & n\sin \theta \\ -n\sin \theta & n\cos \theta \end{array}\right]$
• D. $\left[\begin{array}{cc}-\cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right]$

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

$\Rightarrow A^2=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$

$=\left[\begin{array}{cc}{\cos }^2\theta -{\sin }^2\theta & 2\sin \theta \cos \theta \\ -\sin \theta \cos \theta & {\cos }^2\theta -{\sin }^2\theta \end{array}\right]$

$=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$

Therefore,

$\Rightarrow A^n=A^{n-1}.A$

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

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

Hence, the answer is $\left[\begin{array}{cc}\cos n\theta & \sin n\theta \\ -\sin n\theta & \cos n\theta \end{array}\right].$