Question

If $E\left(\theta \right)=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$ then $E\left(\alpha \right)E\left(\beta \right)$ is equal to

• A. $E\left(\alpha \beta \right)$
• B. $E(\alpha -\beta )$
• C. $E(\alpha +\beta )$
• D. $-E(\alpha +\beta )$

Answer 1

Answer: (C)

$E(\alpha +\beta )$

$E\left(\theta \right)=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]E\left(\alpha \right)E\left(\beta \right)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc}\cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \end{array}\right]⟹\left[\begin{array}{cc}\cos \alpha \cos \beta -\sin \alpha \sin \beta & \cos \alpha \sin \beta +\sin \alpha \cos \beta \\ -\sin \alpha \cos \beta -\cos \alpha \sin \beta & -\sin \alpha \sin \beta +\cos \alpha \cos \beta \end{array}\right]⟹\left[\begin{array}{cc}\cos (\alpha +\beta )& \sin (\alpha +\beta )\\ -\sin (\alpha +\beta )& \cos (\alpha +\beta )\end{array}\right]⟹E(\alpha +\beta )$

Hence, option C is correct.