A) \[\frac{\pi }{16}\]
B) 0
C) \[\frac{\pi }{32}\]
D) \[\frac{\pi }{64}\]
Correct Answer: D
Solution :
\[A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[{{A}^{2}}=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \\ \end{matrix} \right]\] \[{{A}^{3}}=\left[ \begin{matrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \\ \end{matrix} \right]=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] \[=\left[ \begin{matrix} \cos 3\alpha & -\sin 3\alpha \\ \sin 3\alpha & \cos 3\alpha \\ \end{matrix} \right]\] Similarly\[{{A}^{32}}=\left[ \begin{matrix} \cos 32\alpha & -\sin 32\alpha \\ \sin 32\alpha & \cos 32\alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]\] \[\Rightarrow \]\[\cos 32\alpha =0\And \sin 32\alpha =1\] \[\Rightarrow \]\[32\alpha =(4n+1)\frac{\pi }{2},n\in I\] \[\alpha =(4n+1)\frac{\pi }{64},n\in I\] \[\alpha =\frac{\pi }{64}\text{for}\,n=0\]
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