ক) AI-তৈরি নমুনা উত্তর
ধরি, sin − 1 ( 1 3 ) = A \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) = A sin − 1 ( 3 1 ) = A sin A = 1 3 \sin A = \frac{1}{\sqrt{3}} sin A = 3 1 cos 2 ( sin − 1 1 3 ) = cos 2 A = 1 − sin 2 A = 1 − ( 1 3 ) 2 = 1 − 1 3 = 2 3 \cos^2\left(\sin^{-1} \frac{1}{\sqrt{3}}\right) = \cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{1}{\sqrt{3}}\right)^2 = 1 - \frac{1}{3} = \frac{2}{3} cos 2 ( sin − 1 3 1 ) = cos 2 A = 1 − sin 2 A = 1 − ( 3 1 ) 2 = 1 − 3 1 = 3 2
খ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে, f ( θ ) = sin θ f(\theta) = \sin\theta f ( θ ) = sin θ 2 f ( θ ) = 2 sin θ \sqrt{2}f(\theta) = \sqrt{2}\sin\theta 2 f ( θ ) = 2 sin θ f ( π 2 − 2 θ ) = sin ( π 2 − 2 θ ) = cos 2 θ f\left(\frac{\pi}{2} - 2\theta\right) = \sin\left(\frac{\pi}{2} - 2\theta\right) = \cos 2\theta f ( 2 π − 2 θ ) = sin ( 2 π − 2 θ ) = cos 2 θ f ( π 2 − 2 θ ) = cos 2 θ \sqrt{f\left(\frac{\pi}{2} - 2\theta\right)} = \sqrt{\cos 2\theta} f ( 2 π − 2 θ ) = cos 2 θ
বামপক্ষ = sin − 1 ( 2 sin θ ) + sin − 1 ( cos 2 θ ) = \sin^{-1}(\sqrt{2}\sin\theta) + \sin^{-1}(\sqrt{\cos 2\theta}) = sin − 1 ( 2 sin θ ) + sin − 1 ( cos 2 θ ) sin − 1 x + sin − 1 y = sin − 1 ( x 1 − y 2 + y 1 − x 2 ) \sin^{-1} x + \sin^{-1} y = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) sin − 1 x + sin − 1 y = sin − 1 ( x 1 − y 2 + y 1 − x 2 ) x = 2 sin θ ⟹ 1 − x 2 = 1 − 2 sin 2 θ = cos 2 θ x = \sqrt{2}\sin\theta \implies \sqrt{1-x^2} = \sqrt{1 - 2\sin^2\theta} = \sqrt{\cos 2\theta} x = 2 sin θ ⟹ 1 − x 2 = 1 − 2 sin 2 θ = cos 2 θ y = cos 2 θ ⟹ 1 − y 2 = 1 − cos 2 θ = 2 sin 2 θ = 2 sin θ y = \sqrt{\cos 2\theta} \implies \sqrt{1-y^2} = \sqrt{1 - \cos 2\theta} = \sqrt{2\sin^2\theta} = \sqrt{2}\sin\theta y = cos 2 θ ⟹ 1 − y 2 = 1 − cos 2 θ = 2 sin 2 θ = 2 sin θ
অতএব,= sin − 1 [ ( 2 sin θ ) ( 2 sin θ ) + ( cos 2 θ ) ( cos 2 θ ) ] = \sin^{-1}\left[(\sqrt{2}\sin\theta)(\sqrt{2}\sin\theta) + (\sqrt{\cos 2\theta})(\sqrt{\cos 2\theta})\right] = sin − 1 [ ( 2 sin θ ) ( 2 sin θ ) + ( cos 2 θ ) ( cos 2 θ ) ] = sin − 1 ( 2 sin 2 θ + cos 2 θ ) = \sin^{-1}\left(2\sin^2\theta + \cos 2\theta\right) = sin − 1 ( 2 sin 2 θ + cos 2 θ ) = sin − 1 ( 2 sin 2 θ + 1 − 2 sin 2 θ ) = \sin^{-1}\left(2\sin^2\theta + 1 - 2\sin^2\theta\right) = sin − 1 ( 2 sin 2 θ + 1 − 2 sin 2 θ ) = sin − 1 ( 1 ) = π 2 = = \sin^{-1}(1) = \frac{\pi}{2} = = sin − 1 ( 1 ) = 2 π =
গ) AI-তৈরি নমুনা উত্তর
দেওয়া আছে,f ( π 2 − θ ) + 3 f ( θ ) = 2 f\left(\frac{\pi}{2} - \theta\right) + \sqrt{3}f(\theta) = \sqrt{2} f ( 2 π − θ ) + 3 f ( θ ) = 2 sin ( π 2 − θ ) + 3 sin θ = 2 \sin\left(\frac{\pi}{2} - \theta\right) + \sqrt{3}\sin\theta = \sqrt{2} sin ( 2 π − θ ) + 3 sin θ = 2 cos θ + 3 sin θ = 2 \cos\theta + \sqrt{3}\sin\theta = \sqrt{2} cos θ + 3 sin θ = 2
উভয়পক্ষকে 1 2 + ( 3 ) 2 = 1 + 3 = 2 \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2 1 2 + ( 3 ) 2 = 1 + 3 = 2 1 2 cos θ + 3 2 sin θ = 2 2 \frac{1}{2}\cos\theta + \frac{\sqrt{3}}{2}\sin\theta = \frac{\sqrt{2}}{2} 2 1 cos θ + 2 3 sin θ = 2 2 cos θ cos π 3 + sin θ sin π 3 = 1 2 \cos\theta \cos\frac{\pi}{3} + \sin\theta \sin\frac{\pi}{3} = \frac{1}{\sqrt{2}} cos θ cos 3 π + sin θ sin 3 π = 2 1 cos ( θ − π 3 ) = cos π 4 \cos\left(\theta - \frac{\pi}{3}\right) = \cos\frac{\pi}{4} cos ( θ − 3 π ) = cos 4 π
সাধারণ সমাধান,θ − π 3 = 2 n π ± π 4 \theta - \frac{\pi}{3} = 2n\pi \pm \frac{\pi}{4} θ − 3 π = 2 nπ ± 4 π n ∈ Z n \in \mathbb{Z} n ∈ Z θ = 2 n π + π 3 ± π 4 \theta = 2n\pi + \frac{\pi}{3} \pm \frac{\pi}{4} θ = 2 nπ + 3 π ± 4 π
ধনার্থক চিহ্ন নিয়ে: θ = 2 n π + π 3 + π 4 = 2 n π + 7 π 12 \theta = 2n\pi + \frac{\pi}{3} + \frac{\pi}{4} = 2n\pi + \frac{7\pi}{12} θ = 2 nπ + 3 π + 4 π = 2 nπ + 12 7 π θ = 2 n π + π 3 − π 4 = 2 n π + π 12 \theta = 2n\pi + \frac{\pi}{3} - \frac{\pi}{4} = 2n\pi + \frac{\pi}{12} θ = 2 nπ + 3 π − 4 π = 2 nπ + 12 π
অতএব, নির্ণেয় সমাধান: θ = 2 n π + 7 π 12 \theta = 2n\pi + \frac{7\pi}{12} θ = 2 nπ + 12 7 π 2 n π + π 12 2n\pi + \frac{\pi}{12} 2 nπ + 12 π n ∈ Z n \in \mathbb{Z} n ∈ Z