JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 6)
Let $$A = \left( {\matrix{
{\cos \alpha } & { - \sin \alpha } \cr
{\sin \alpha } & {\cos \alpha } \cr
} } \right)$$, ($$\alpha $$ $$ \in $$ R)
such that $${A^{32}} = \left( {\matrix{ 0 & { - 1} \cr 1 & 0 \cr } } \right)$$ then a value of $$\alpha $$ is
such that $${A^{32}} = \left( {\matrix{ 0 & { - 1} \cr 1 & 0 \cr } } \right)$$ then a value of $$\alpha $$ is
0
$${\pi \over {16}}$$
$${\pi \over {32}}$$
$${\pi \over {64}}$$
Explanation
_8th_April_Morning_Slot_en_6_2.png)
From here sin 32$$\alpha $$ = 1 and cos 32$$\alpha $$ = 0
$$ \therefore $$ 32$$\alpha $$ = 2n$$\pi $$ + $${\pi \over 2}$$
$$ \Rightarrow $$ $$\alpha $$ = $${\pi \over {64}} + {{n\pi } \over {16}}$$
Putting n = 0, $$\alpha $$ = $${\pi \over {64}}$$
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