A) \[0\]
B) \[\frac{\pi }{32}\]
C) \[\frac{\pi }{64}\]
D) \[\frac{\pi }{16}\]
Correct Answer: C
Solution :
| \[{{A}^{2}}=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\] |
| \[{{A}^{2}}=\left[ \begin{matrix} \cos 2\alpha & -\sin 2\alpha \\ \sin 2\alpha & \cos 2\alpha \\ \end{matrix} \right]\] |
| Similarly we observe that, |
| \[{{A}^{n}}=\left[ \begin{matrix} \operatorname{cosn}\alpha & -\operatorname{sinn}\alpha \\ \operatorname{sinn}\alpha & \operatorname{cosn}\alpha \\ \end{matrix} \right]\] |
| Hence,\[\cos 32\alpha =0\]and\[sin32\alpha =1\] |
| \[32\alpha =2n\pi +\frac{\pi }{2}\] |
| \[\alpha =\frac{n\pi }{16}+\frac{\pi }{64},n\in i.\] |
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