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 ]
তাহলে,A 2 = [ cos θ − sin θ sin θ cos θ ] [ cos θ − sin θ sin θ cos θ ] A^2 = \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 = [ cos θ sin θ − sin θ cos θ ] [ cos θ sin θ − sin θ cos θ ] = [ cos 2 θ − sin 2 θ − 2 sin θ cos θ 2 sin θ cos θ cos 2 θ − sin 2 θ ] = [ cos 2 θ − sin 2 θ sin 2 θ cos 2 θ ] = \begin{bmatrix} \cos^2\theta - \sin^2\theta & -2\sin\theta\cos\theta \\ 2\sin\theta\cos\theta & \cos^2\theta - \sin^2\theta \end{bmatrix} = \begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{bmatrix} = [ cos 2 θ − sin 2 θ 2 sin θ cos θ − 2 sin θ cos θ cos 2 θ − sin 2 θ ] = [ cos 2 θ sin 2 θ − sin 2 θ cos 2 θ ]
দেওয়া আছে,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 ]
ম্যাট্রিক্সদ্বয়ের সমতা হতে পাই,cos 2 θ = 1 2 এবং sin 2 θ = 3 2 \cos 2\theta = \frac{1}{2} \quad \text{এবং} \quad \sin 2\theta = \frac{\sqrt{3}}{2} cos 2 θ = 2 1 এবং sin 2 θ = 2 3
অতএব,2 θ = 2 n π + π 3 ⟹ θ = n π + π 6 ( n ∈ Z ) 2\theta = 2n\pi + \frac{\pi}{3} \implies \theta = n\pi + \frac{\pi}{6} \quad (n \in \mathbb{Z}) 2 θ = 2 nπ + 3 π ⟹ θ = nπ + 6 π ( n ∈ Z )
প্রধান মান হিসেবে, θ = π 6 \theta = \frac{\pi}{6} θ = 6 π 30 ∘ 30^\circ 3 0 ∘