JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 9)
If $$A = \left[ {\matrix{
{\cos \theta } & {i\sin \theta } \cr
{i\sin \theta } & {\cos \theta } \cr
} } \right]$$, $$\left( {\theta = {\pi \over {24}}} \right)$$
and $${A^5} = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$$, where $$i = \sqrt { - 1} $$ then which one of the following is not true?
and $${A^5} = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$$, where $$i = \sqrt { - 1} $$ then which one of the following is not true?
$$a$$2 - $$c$$2 = 1
$$0 \le {a^2} + {b^2} \le 1$$
$$ a$$2 - $$d$$2 = 0
$${a^2} - {b^2} = {1 \over 2}$$
Explanation
$$ \because $$ $$A = \left[ {\matrix{
{\cos \theta } & {i\sin \theta } \cr
{i\sin \theta } & {\cos \theta } \cr
} } \right]$$
$$ \therefore $$ $${A^n} = \left[ {\matrix{ {\cos \,n\theta } & {i\sin \,n\theta } \cr {i\sin \,n\theta } & {\cos \,n\theta } \cr } } \right],n \in N$$
$$ \therefore $$$${A^5} = \left[ {\matrix{ {\cos \,5\theta } & {i\sin \,5\theta } \cr {i\sin \,5\theta } & {\cos \,5\theta } \cr } } \right] = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$$
$$ \therefore $$ $$a = \cos 5\theta ,\,b = i\sin 5\theta = c,\,d = \cos 5\theta $$
$$ \therefore $$ $${a^2} - {b^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$$
$${a^2} - {c^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$$
$${a^2} - {d^2} = {\cos ^2}5\theta - {\cos ^2}5\theta = 1$$
$${a^2} + {b^2} = {\cos ^2}5\theta - {\sin ^2}5\theta = \cos 10\theta = \cos {{10\pi } \over {24}}$$
and $$0 < \cos {{5\pi } \over {12}} < 1 $$
$$\Rightarrow 0 \le {a^2} + {b^2} \le 1$$
$$ \therefore $$ $${A^n} = \left[ {\matrix{ {\cos \,n\theta } & {i\sin \,n\theta } \cr {i\sin \,n\theta } & {\cos \,n\theta } \cr } } \right],n \in N$$
$$ \therefore $$$${A^5} = \left[ {\matrix{ {\cos \,5\theta } & {i\sin \,5\theta } \cr {i\sin \,5\theta } & {\cos \,5\theta } \cr } } \right] = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$$
$$ \therefore $$ $$a = \cos 5\theta ,\,b = i\sin 5\theta = c,\,d = \cos 5\theta $$
$$ \therefore $$ $${a^2} - {b^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$$
$${a^2} - {c^2} = {\cos ^2}5\theta + {\sin ^2}5\theta = 1$$
$${a^2} - {d^2} = {\cos ^2}5\theta - {\cos ^2}5\theta = 1$$
$${a^2} + {b^2} = {\cos ^2}5\theta - {\sin ^2}5\theta = \cos 10\theta = \cos {{10\pi } \over {24}}$$
and $$0 < \cos {{5\pi } \over {12}} < 1 $$
$$\Rightarrow 0 \le {a^2} + {b^2} \le 1$$
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