Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If [cos theta sin theta ] then show that A^2 equals [cos2theta sin2theta]

If A=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right], then show that \quad \mathrm{A}^{2}=\left[\begin{array}{cc} \cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta \end{array}\right]

Answers (1)

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

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

According to the rule of matrix multiplication:
\\$$ \mathrm{A}^{2}=\left[\begin{array}{cc} \cos \theta \times \cos \theta+\sin \theta \times(-\sin \theta) & \cos \theta \times \sin \theta+\sin \theta \times \cos \theta \\ -\cos \theta \times \sin \theta+(-\sin \theta \times \cos \theta) & \cos \theta \times \cos \theta+\sin \theta \times(-\sin \theta) \end{array}\right] $$ \\ \Rightarrow \mathrm{A}^{2}=\left[\begin{array}{cc}\cos ^{2} \theta-\sin ^{2} \theta & 2 \sin \theta \cos \theta \\ -2 \sin \theta \cos \theta & \cos ^{2} \theta-\sin ^{2} \theta\end{array}\right]
We know that:
\\2 \sin \theta \cos \theta=\sin 2 \theta$ and $\cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta \\\therefore A^{2}=\left[\begin{array}{cc}\cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta\end{array}\right]_{\ldots}
Hence.proved.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question