Prove the following

Question:

If $A=\left[\begin{array}{cc}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right],\left(\theta=\frac{\pi}{24}\right)$ and $A^{5}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$,

where $i=\sqrt{-1}$, then which one of the following is not true?

 

  1. (1) $0 \leq a^{2}+b^{2} \leq 1$

  2. (2) $a^{2}-d^{2}=0$

  3. (3) $a^{2}-c^{2}=1$

  4. (4) $a^{2}-b^{2}=\frac{1}{2}$


Correct Option: , 4

Solution:

$\because A=\left[\begin{array}{cc}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right]$

$\therefore A^{n}=\left[\begin{array}{cc}\cos n \theta & i \sin n \theta \\ i \sin n \theta & \cos n \theta\end{array}\right], n \in \mathbf{N}$

$\because A^{5}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$\therefore A^{5}=\left[\begin{array}{cc}\cos 5 \theta & i \sin 5 \theta \\ i \sin 5 \theta & \cos 5 \theta\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$

$\therefore a=\cos 5 \theta, b=i \sin 5 \theta=c, d=\cos 5 \theta$

$\therefore a^{2}-b^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1$

$a^{2}-c^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1$

$a^{2}-d^{2}=\cos ^{2} 5 \theta-\cos ^{2} 5 \theta=1$

$a^{2}+b^{2}=\cos ^{2} 5 \theta-\sin ^{2} 5 \theta=\cos 10 \theta=\cos \frac{10 \pi}{24}$

and $0<\cos \frac{5 \pi}{12}<1 \Rightarrow 0 \leq a^{2}+b^{2} \leq 1$

$\therefore a^{2}-b^{2}=\frac{1}{2}$ is wrong.

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