যদি 0°<θ<180° হয়, তবে  sqrt(2+sqrt(2+sqrt(........+2(1+costheta))))=? 

প্রশ্ন: যদি \(0° < \theta < 180°\) হয়, তবে \(\sqrt{2+\sqrt{2+\sqrt{........+2(1+\cos\theta)}}} = ?\)

উত্তর: \(2\cos(\frac{\theta}{2^n})\)

ব্যাখ্যা:

আমরা জানি, \(1 + \cos\theta = 2\cos^2(\frac{\theta}{2})\).

সুতরাং, \(2(1 + \cos\theta) = 4\cos^2(\frac{\theta}{2})\).

এখন, \(\sqrt{2(1 + \cos\theta)} = \sqrt{4\cos^2(\frac{\theta}{2})} = 2\cos(\frac{\theta}{2})\) (যেহেতু \(0° < \theta < 180°\), তাই \(\cos(\frac{\theta}{2})\) ধনাত্মক).

আবার, \(2 + \sqrt{2(1 + \cos\theta)} = 2 + 2\cos(\frac{\theta}{2}) = 2(1 + \cos(\frac{\theta}{2})) = 2 \cdot 2\cos^2(\frac{\theta}{2^2}) = 4\cos^2(\frac{\theta}{2^2})\).

সুতরাং, \(\sqrt{2 + \sqrt{2(1 + \cos\theta)}} = \sqrt{4\cos^2(\frac{\theta}{2^2})} = 2\cos(\frac{\theta}{2^2})\).

একইভাবে, যদি \(n\) সংখ্যক পদ থাকে, তবে আমরা লিখতে পারি:

\(\sqrt{2+\sqrt{2+\sqrt{........+2(1+\cos\theta)}}} = 2\cos(\frac{\theta}{2^n})\). 😃

অতএব, উত্তর: \(2\cos(\frac{\theta}{2^n})\). 🎉