AI-তৈরি নমুনা উত্তর
দেওয়া আছে,A = [ cos θ − sin θ sin θ cos θ ] A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} A = [ cos θ sin θ − sin θ cos θ ]
অতএব,A 2 = A ⋅ A = [ cos θ − sin θ sin θ cos θ ] [ cos θ − sin θ sin θ cos θ ] A^2 = A \cdot A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} A 2 = A ⋅ A = [ cos θ sin θ − sin θ cos θ ] [ cos θ sin θ − sin θ cos θ ] = [ cos 2 θ − sin 2 θ − cos θ sin θ − sin θ cos θ sin θ cos θ + cos θ sin θ − sin 2 θ + cos 2 θ ] = \begin{bmatrix} \cos^2 \theta - \sin^2 \theta & -\cos \theta \sin \theta - \sin \theta \cos \theta \\ \sin \theta \cos \theta + \cos \theta \sin \theta & -\sin^2 \theta + \cos^2 \theta \end{bmatrix} = [ cos 2 θ − sin 2 θ sin θ cos θ + cos θ sin θ − cos θ sin θ − sin θ cos θ − sin 2 θ + cos 2 θ ] = [ cos 2 θ − sin 2 θ sin 2 θ cos 2 θ ] = \begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{bmatrix} = [ cos 2 θ sin 2 θ − sin 2 θ cos 2 θ ]
আবার প্রশ্নমতে,A 2 = 1 2 [ 1 − 3 3 1 ] = [ 1 2 − 3 2 3 2 1 2 ] A^2 = \frac{1}{2}\begin{bmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} A 2 = 2 1 [ 1 3 − 3 1 ] = [ 2 1 2 3 − 2 3 2 1 ]
ম্যাট্রিক্সের সমতা অনুসারে,cos 2 θ = 1 2 \cos 2\theta = \frac{1}{2} cos 2 θ = 2 1 sin 2 θ = 3 2 \sin 2\theta = \frac{\sqrt{3}}{2} sin 2 θ = 2 3
আমরা জানি, cos π 3 = 1 2 \cos \frac{\pi}{3} = \frac{1}{2} cos 3 π = 2 1 sin π 3 = 3 2 \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} sin 3 π = 2 3 2 θ = 2 n π + π 3 , n ∈ Z 2\theta = 2n\pi + \frac{\pi}{3}, \quad n \in \mathbb{Z} 2 θ = 2 nπ + 3 π , n ∈ Z ⟹ θ = n π + π 6 , n ∈ Z \implies \theta = n\pi + \frac{\pi}{6}, \quad n \in \mathbb{Z} ⟹ θ = nπ + 6 π , n ∈ Z
মুখ্য মান (Principal value) বিবেচনায়,θ = π 6 বা 30 ∘ \theta = \frac{\pi}{6} \text{ বা } 30^\circ θ = 6 π বা 3 0 ∘
উত্তর: θ = n π + π 6 , n ∈ Z \theta = n\pi + \frac{\pi}{6}, \; n \in \mathbb{Z} θ = nπ + 6 π , n ∈ Z θ = π 6 \theta = \frac{\pi}{6} θ = 6 π