If A = begin{bmatrix} cos theta & i sin theta | vedclass

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(C) Given $A = \begin{bmatrix} \cos 	heta & i \sin 	heta \ i \sin 	heta & \cos 	heta \end{bmatrix}$.Using the property of this specific matrix fo

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$a^{2} - b^{2} = \frac{1}{2}$

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(C) Given $A = \begin{bmatrix} \cos 	heta & i \sin 	heta \ i \sin 	heta & \cos 	heta \end{bmatrix}$.Using the property of this specific matrix form,$A^{n} = \begin{bmatrix} \cos n	heta & i \sin n	heta \ i \sin n	heta & \cos n	heta \end{bmatrix}$.For $n = 5$,$A^{5} = \begin{bmatrix} \cos 5	heta & i \sin 5	heta \ i \sin 5	heta & \cos 5	heta \end{bmatrix} = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Thus,$a = \cos 5	heta$,$b = i \sin 5	heta$,$c = i \sin 5	heta$,$d = \cos 5	heta$.Now,let us evaluate the options:$a^{2} + b^{2} = \cos^{2} 5	heta + (i \sin 5	heta)^{2} = \cos^{2} 5	heta - \sin^{2} 5	heta = \cos 10	heta = \cos(10 	imes \frac{\pi}{24}) = \cos(\frac{5\pi}{12}) = \cos 75^{\circ}$. Since $0 \leq \cos 75^{\circ} \leq 1$,option $A$ is true.$a^{2} - d^{2} = \cos^{2} 5	heta - \cos^{2} 5	heta = 0$. Option $B$ is true.$a^{2} - b^{2} = \cos^{2} 5	heta - (i \sin 5	heta)^{2} = \cos^{2} 5	heta + \sin^{2} 5	heta = 1$. Option $C$ states $a^{2} - b^{2} = \frac{1}{2}$,which is false.$a^{2} - c^{2} = \cos^{2} 5	heta - (i \sin 5	heta)^{2} = \cos^{2} 5	heta + \sin^{2} 5	heta = 1$. Option $D$ is true.Therefore,the statement that is not true is $a^{2} - b^{2} = \frac{1}{2}$.

If $A = \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix}$,$\theta = \frac{\pi}{24}$ and $A^{5} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,where $i = \sqrt{-1}$,then which one of the following is not true?

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