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. $E\left(0^o\right)$
• B. $E\left(\alpha \beta \right)$
• C. $E(\alpha +\beta )$
• D. $E(\alpha -\beta )$

Answer 1

The correct option is C $E(\alpha +\beta )$

Given,

$E\left(\theta \right)=\left(\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right)$

$E\left(\alpha \right).E\left(\beta \right)=\left(\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right)\left(\begin{array}{cc}cos\beta & sin\beta \\ -sin\beta & cos\beta \end{array}\right)$

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

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

$=E(\alpha +\beta )$

Hence $E\left(\alpha \right)E\left(\beta \right)=E(\alpha +\beta )$