দৃশ্যকল্প: f($\theta$) = cos$\theta$.

Dhaka • 2025

ক) প্রমাণ কর: $\frac{1}{2} \cos^{-1}x = \cos^{-1} \sqrt{\frac{1+x}{2}}$ ।

খ) P = f($\frac{\pi}{2} - \theta$) এবং Q = f(2$\theta$) হলে প্রমাণ কর যে, $\sin^{-1}(\sqrt{2}P) + \sin^{-1}\sqrt{Q} = \frac{\pi}{2}$ ।

গ) সমাধান কর: 4f($\theta$)f(2$\theta$)f(3$\theta$) = 1, যখন 0° < $\theta$ < $\pi$ ।

ক-এর উত্তর:
ধরি, $\cos^{-1} x = \theta$
$\therefore x = \cos \theta$

ডানপক্ষ = $\cos^{-1} \sqrt{\frac{1+x}{2}}$
$= \cos^{-1} \sqrt{\frac{1+\cos \theta}{2}}$
$= \cos^{-1} \sqrt{\frac{2 \cos^{2} (\theta/2)}{2}}$ [যেহেতু $1+\cos \theta = 2 \cos^{2} (\theta/2)$]
$= \cos^{-1} (\cos \frac{\theta}{2})$
$= \frac{1}{2} \theta$
$= \frac{1}{2} \cos^{-1} x$ = বামপক্ষ (প্রমাণিত)

খ-এর উত্তর:
দেওয়া আছে, $f(\theta) = \cos \theta$
$\therefore P = f(\frac{\pi}{2} - \theta) = \cos(\frac{\pi}{2} - \theta) = \sin \theta$
এবং $Q = f(2\theta) = \cos 2\theta$

বামপক্ষ = $\sin^{-1} (\sqrt{2}P) + \sin^{-1} \sqrt{Q}$
$= \sin^{-1} (\sqrt{2} \sin \theta) + \sin^{-1} \sqrt{\cos 2\theta}$
ধরি, $\sin^{-1} (\sqrt{2} \sin \theta) = A \implies \sin A = \sqrt{2} \sin \theta \implies \sin^{2} A = 2 \sin^{2} \theta$
এবং $\sin^{-1} \sqrt{\cos 2\theta} = B \implies \sin B = \sqrt{\cos 2\theta} \implies \sin^{2} B = \cos 2\theta$

এখন, $\sin^{2} B = 1 - 2 \sin^{2} \theta$
বা, $\sin^{2} B = 1 - \sin^{2} A$
বা, $\sin^{2} A + \sin^{2} B = 1$
বা, $\sin^{2} A = 1 - \sin^{2} B = \cos^{2} B$
$\therefore \sin A = \cos B = \sin (\frac{\pi}{2} - B)$
বা, $A = \frac{\pi}{2} - B \implies A + B = \frac{\pi}{2}$
অর্থাৎ, $\sin^{-1} (\sqrt{2}P) + \sin^{-1} \sqrt{Q} = \frac{\pi}{2}$ (প্রমাণিত)

গ-এর উত্তর:
প্রদত্ত সমীকরণ: $4 \cos \theta \cdot \cos 2\theta \cdot \cos 3\theta = 1$
বা, $2 \cos 2\theta \cdot (2 \cos 3\theta \cdot \cos \theta) = 1$
বা, $2 \cos 2\theta (\cos 4\theta + \cos 2\theta) = 1$
বা, $2 \cos 4\theta \cdot \cos 2\theta + 2 \cos^{2} 2\theta = 1$
বা, $\cos 6\theta + \cos 2\theta + 1 + \cos 4\theta = 1$
বা, $\cos 6\theta + \cos 4\theta + \cos 2\theta = 0$
বা, $(\cos 6\theta + \cos 2\theta) + \cos 4\theta = 0$
বা, $2 \cos 4\theta \cdot \cos 2\theta + \cos 4\theta = 0$
বা, $\cos 4\theta (2 \cos 2\theta + 1) = 0$

হয়, $\cos 4\theta = 0 \implies 4\theta = (2n+1)\frac{\pi}{2} \implies \theta = (2n+1)\frac{\pi}{8}$
$n=0, 1, 2, 3$ হলে, $\theta = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}$ (সীমার মধ্যে)

অথবা, $2 \cos 2\theta + 1 = 0 \implies \cos 2\theta = -\frac{1}{2} = \cos \frac{2\pi}{3}$
$\therefore 2\theta = 2n\pi \pm \frac{2\pi}{3} \implies \theta = n\pi \pm \frac{\pi}{3}$
$n=0, 1$ হলে, $\theta = \frac{\pi}{3}, \frac{2\pi}{3}$ (সীমার মধ্যে)

$\therefore \theta = \{\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{\pi}{3}, \frac{2\pi}{3}\}$

সমাধান (Solution)