Question

If \[ A(\alpha)= \begin{pmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{pmatrix} \] then $A(\alpha)\,A(\beta)=$

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

Hint:

Rotation matrices follow angle addition: multiplying two rotation matrices adds their angles.

Answer 1

The Correct Option is C

Step 1: Write both matrices: \[ A(\alpha)= \begin{pmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{pmatrix},\quad A(\beta)= \begin{pmatrix} \cos\beta & \sin\beta\\ -\sin\beta & \cos\beta \end{pmatrix} \]
Step 2: Multiply $A(\alpha)$ and $A(\beta)$: \[ A(\alpha)A(\beta)= \begin{pmatrix} \cos\alpha\cos\beta-\sin\alpha\sin\beta & \cos\alpha\sin\beta+\sin\alpha\cos\beta\\ -(\sin\alpha\cos\beta+\cos\alpha\sin\beta) & \cos\alpha\cos\beta-\sin\alpha\sin\beta \end{pmatrix} \]
Step 3: Use trigonometric identities: \[ \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta \] \[ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta \]
Step 4: Substitute: \[ A(\alpha)A(\beta)= \begin{pmatrix} \cos(\alpha+\beta) & \sin(\alpha+\beta)\\ -\sin(\alpha+\beta) & \cos(\alpha+\beta) \end{pmatrix} \]
Step 5: This is exactly $A(\alpha+\beta)$.