Simplify the following............

1.	Simplify the following............
Answer» QUESTION: SIMPLIFY: $cos \ \theta\left[\begin{array}{cc}cos \ \theta&sin \ \theta\\ \\ -sin\ \theta&cos \ \theta\end{array}\right] \\ + sin \ \theta\left[\begin{array}{cc}sin \ \theta&-cos \ \theta\\ \\ cos\ \theta&sin \ \theta\end{array}\right]$ ANSWER: $\left[\begin{array}{cc}1&0\\ \\0&1\end{array}\right]$ GIVEN: $cos \ \theta\left[\begin{array}{cc}cos \ \theta&sin \ \theta\\ \\ -sin\ \theta&cos \ \theta\end{array}\right] \\ + sin \ \theta\left[\begin{array}{cc}sin \ \theta&-cos \ \theta\\ \\ cos\ \theta&sin \ \theta\end{array}\right]$ TO FIND: $cos \ \theta\left[\begin{array}{cc}cos \ \theta&sin \ \theta\\ \\ -sin\ \theta&cos \ \theta\end{array}\right] \\ + sin \ \theta\left[\begin{array}{cc}sin \ \theta&-cos \ \theta\\ \\ cos\ \theta&sin \ \theta\end{array}\right]$ EXPLANATION: $cos \ \theta\left[\begin{array}{cc}cos \ \theta&sin \ \theta\\ \\ -sin\ \theta&cos \ \theta\end{array}\right] \\ + sin \ \theta\left[\begin{array}{cc}sin \ \theta&-cos \ \theta\\ \\ cos\ \theta&sin \ \theta\end{array}\right]$ $Let \ A = \left[\begin{array}{cc}cos \ \theta&sin \ \theta\\ \\ -sin\ \theta&cos \ \theta\end{array}\right]$ $Let \ B =\left[\begin{array}{cc}sin \ \theta&-cos \ \theta\\ \\ cos\ \theta&sin \ \theta\end{array}\right]$ $cos \ \theta \times A=\left[\begin{array}{cc}cos^2 \ \theta&cos \ \theta \ sin \ \theta\\ \\ -sin\ \theta\ cos \ \theta&cos^2 \ \theta\end{array}\right]$ $sin \ \theta\times B = \left[\begin{array}{cc}sin^2 \ \theta&-sin \ \theta\ cos \ \theta\\ \\ cos\ \theta \ sin \ \theta&sin^2 \ \theta\end{array}\right]$ $Let\ cos \ \theta \times A+sin \ \theta \times B =\left[\begin{array}{cc}a_{11}&a_{12}\\ \\ a_{21}&a_{22}\end{array}\right]$ $a_{11}= sin^2\ \theta + cos^2\ \theta$ $a_{12}= sin \ \theta \ cos\ \theta-sin \ \theta \ cos\ \theta$ $a_{21}= sin \ \theta \ cos\ \theta-sin \ \theta \ cos\ \theta$ $a_{22}= sin^2\ \theta + cos^2\ \theta$ $\boxed{\bold{\large{sin^2 \ \theta + cos^2 \ \theta = 1}}}$ $a_{11} = 1$ $a_{12}=0$ $a_{21}=0$ $a_{22}= 1$ Substituting these values $\left[\begin{array}{cc}1&0\\ \\0&1\end{array}\right]$ $\left[\begin{array}{cc}1&0\\ \\0&1\end{array}\right]= I$ $cos \ \theta\left[\begin{array}{cc}cos \ \theta&sin \ \theta\\ \\ -sin\ \theta&cos \ \theta\end{array}\right]\\ + sin \ \theta\left[\begin{array}{cc}sin \ \theta&-cos \ \theta\\ \\ cos\ \theta&sin \ \theta\end{array}\right]= I$

Answer»

QUESTION:

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

ANSWER:

$\left[\begin{array}{cc}1&0\\ \\0&1\end{array}\right]$

GIVEN:

TO FIND:

EXPLANATION:

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

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

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

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

$Let\ cos \ \theta \times A+sin \ \theta \times B =\left[\begin{array}{cc}a_{11}&a_{12}\\ \\ a_{21}&a_{22}\end{array}\right]$

$a_{11}= sin^2\ \theta + cos^2\ \theta$

$a_{12}= sin \ \theta \ cos\ \theta-sin \ \theta \ cos\ \theta$

$a_{21}= sin \ \theta \ cos\ \theta-sin \ \theta \ cos\ \theta$

$a_{22}= sin^2\ \theta + cos^2\ \theta$

$\boxed{\bold{\large{sin^2 \ \theta + cos^2 \ \theta = 1}}}$

$a_{11} = 1$

$a_{12}=0$

$a_{21}=0$

$a_{22}= 1$

Substituting these values

$\left[\begin{array}{cc}1&0\\ \\0&1\end{array}\right]$

$\left[\begin{array}{cc}1&0\\ \\0&1\end{array}\right]= I$

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