# Solve this

Question:

If $\mathrm{A}=\left[\begin{array}{cc}\cos \theta & \text { isin } \theta \\ i \sin \theta & \cos \theta\end{array}\right], \quad\left(\theta=\frac{\pi}{24}\right)$ and

$\mathrm{A}^{5}=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]$, where $\mathrm{i}=\sqrt{-1}$, then which one

of the following is not true?

1. 0 \leq a^{2}+b^{2} \leq 1$2.$\mathrm{a}^{2}-\mathrm{d}^{2}=0$3.$a^{2}-b^{2}=\frac{1}{2}$4.$a^{2}-c^{2}=1$Correct Option: , 3 Solution:$A^{2}=\left(\begin{array}{cc}\cos 2 \theta & i \sin 2 \theta \\ i \sin 2 \theta & \cos 2 \theta\end{array}\right)$Similarly,$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)$(1)$a^{2}+b^{2}=\cos ^{2} 5 \theta-\sin ^{2} 5 \theta=\cos 10 \theta=\cos 75^{\circ}$(2)$a^{2}-d^{2}=\cos ^{2} 5 \theta-\cos ^{2} 5 \theta=0$(3)$a^{2}-b^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1$(4)$a^{2}-c^{2}=\cos ^{2} 5 \theta+\sin ^{2} 5 \theta=1\$