Question

Evaluate ${\left(\begin{array}{cc}\cos \theta +\sin \theta & \sqrt{2}\sin \theta \\ -\sqrt{2}\sin \theta & \cos \theta -\sin \theta \end{array}\right)}^n$

• A. $\left(\begin{array}{cc}-\cos n\theta +\sin n\theta & \sqrt{2}\sin n\theta \\ \sqrt{2}\sin n\theta & \cos n\theta -\sin n\theta \end{array}\right)$
• B. $\left(\begin{array}{cc}\cos n\theta +\sin n\theta & \sqrt{2}\sin n\theta \\ -\sqrt{2}\sin n\theta & \cos n\theta -\sin n\theta \end{array}\right)$
• C. $\left(\begin{array}{cc}-\cos n\theta +-\sin n\theta & \sqrt{2}\sin n\theta \\ \sqrt{2}\sin n\theta & \cos n\theta -\sin n\theta \end{array}\right)$
• D. $\left(\begin{array}{cc}\cos n\theta +-\sin n\theta & \sqrt{2}\sin n\theta \\ \sqrt{2}\sin n\theta & \cos n\theta +\sin n\theta \end{array}\right)$

Answer 1

Answer: (B)

$\left(\begin{array}{cc}\cos n\theta +\sin n\theta & \sqrt{2}\sin n\theta \\ -\sqrt{2}\sin n\theta & \cos n\theta -\sin n\theta \end{array}\right)$

$A^1=\left[\begin{array}{cc}\cos \theta +\sin \theta & \sqrt{2}\sin \theta \\ -\sqrt{2}\sin \theta & \cos \theta -\sin \theta \end{array}\right]A^2=\left[\begin{array}{cc}\cos \theta +\sin \theta & \sqrt{2}\sin \theta \\ -\sqrt{2}\sin \theta & \cos \theta -\sin \theta \end{array}\right]\left[\begin{array}{cc}\cos \theta +\sin \theta & \sqrt{2}\sin \theta \\ -\sqrt{2}\sin \theta & \cos \theta -\sin \theta \end{array}\right]=\left[\begin{array}{cc}1+\sin 2\theta -2{\sin }^2\theta & \sqrt{2}\sin \theta \cos \theta +\sqrt{2}\sin \theta \cos \theta +\sqrt{2}\sin \theta \cos \theta \\ -\sqrt{2}\sin \theta \cos \theta -\sqrt{2}\sin \theta \cos \theta & -2{\sin }^2\theta +1-\sin 2\theta \end{array}\right]\therefore A^2\left[\begin{array}{cc}\sin 2\theta +\cos 2\theta & \sqrt{2}\sin 2\theta \\ -\sqrt{2}\sin 2\theta & \cos 2\theta -\sin 2\theta \end{array}\right]A^n=\left[\begin{array}{cc}\sin n\theta +\cos n\theta & \sqrt{2}\sin n\theta \\ -\sqrt{2}\sin n\theta & \cos n\theta -\sin n\theta \end{array}\right]$