Math, asked by Anonymous, 1 month ago


Simplify

[tex] \cos \theta\left[
\begin{array}{c c c}
\cos \theta& \sin \theta \\
& & \\
- \sin \theta & \cos \theta &
\end{array}
\right] + \sin \theta\left[
\begin{array}{c c c}
\sin \theta& - \cos\theta \\
& & \\
\cos\theta & \sin\theta &
\end{array}
\right] [/tex]

Answers

Answered by Anonymous
42

Appropriate Question:-

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

Given Matrix :-

\quad \leadsto \quad \cos\theta\left[\begin{array}{c c c} \cos \theta& \sin \theta \\ & & \\ - \sin \theta & \cos \theta &\end{array}\right] + \sin \theta\left[\begin{array}{c c c} \sin \theta& - \cos\theta \\ & & \\ \cos\theta & \sin\theta & \end{array}\right]

Solution :-

As \sin\theta & \cos\theta are out of matrix . So on , multiplying them with all elements of their respective matrixes , we have ;

 { : \implies \quad \left[\begin{array}{c c c} \cos² \theta& \sin \theta . \cos \theta \\ & & \\ - \sin \theta.  \cos \theta & \cos² \theta &\end{array}\right] + \left[\begin{array}{c c c} \sin² \theta& - \cos\theta . \sin \theta \\ & & \\ \cos\theta .\sin\theta & \sin²\theta & \end{array}\right] }

Now as the order of both matrixes are same i.e 2 × 2 , so we can add their corresponding elements . On adding their corresponding elements we have ;

 { : \implies \quad \left[\begin{array}{c c c} \cos² \theta + \sin²\theta & \sin \theta . \cos \theta - \cos\theta . \sin \theta \\ & & \\ - \sin \theta.  \cos \theta + \sin\theta.  \cos\theta & \cos² \theta + \sin²\theta &\end{array}\right] }

We knows a Trigonometric identity i.e ;

 \quad \qquad { \bigstar { \underline { \boxed { \bf { \red { \cos²\theta + \sin²\theta = 1}}}}}}{\bigstar}

Using this we have ;

 { : \implies \quad \left[\begin{array}{c c} 1 & \cancel{\sin \theta . \cos \theta} - \cancel{\cos\theta . \sin \theta} \\ & & \\ - \cancel{\sin \theta.  \cos \theta} + \cancel{\sin\theta.  \cos\theta} & 1 \end{array}\right] }

 { : \implies \quad \left[\begin{array}{c c} 1 & 0 \\ & \\ 0 & 1 \end{array}\right]}

{ \bigstar { \underline { \boxed { \bf { \red { \therefore\cos\theta\left[\begin{array}{c c c} \cos \theta& \sin \theta \\ & & \\ - \sin \theta & \cos \theta &\end{array}\right] + \sin \theta\left[\begin{array}{c c c} \sin \theta& - \cos\theta \\ & & \\ \cos\theta & \sin\theta & \end{array}\right] = \left[\begin{array}{c c} 1 & 0 \\ & \\ 0 & 1 \end{array}\right] }}}}}}{\bigstar}

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