4cos^2(pi/16)=? 

Explanation:

Another Explanation (5): প্রশ্ন: \(4\cos^2\left(\frac{\pi}{16}\right) = ?\) সমাধান: আমরা জানি, \( \cos(2\theta) = 2\cos^2(\theta) - 1 \)। সুতরাং, \( 2\cos^2(\theta) = 1 + \cos(2\theta) \)। অতএব, \( 4\cos^2(\theta) = 2(1 + \cos(2\theta)) \)। এখানে, \(\theta = \frac{\pi}{16}\)। সুতরাং, \(4\cos^2\left(\frac{\pi}{16}\right) = 2\left(1 + \cos\left(\frac{2\pi}{16}\right)\right) = 2\left(1 + \cos\left(\frac{\pi}{8}\right)\right)\) আবার, \( \cos\left(\frac{\pi}{8}\right) = \cos\left(\frac{\pi/4}{2}\right) \) আমরা জানি, \( \cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \cos(x)}{2}} \) সুতরাং, \( \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \cos(\pi/4)}{2}} = \sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}} = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2} \) এখন, \( 4\cos^2\left(\frac{\pi}{16}\right) = 2\left(1 + \frac{\sqrt{2 + \sqrt{2}}}{2}\right) = 2 + \sqrt{2 + \sqrt{2}} \) এখন আমরা \( \cos(\pi/16) \) বের করার চেষ্টা করি। আমরা জানি, \( \cos(\frac{x}{2}) = \sqrt{\frac{1+\cos x}{2}} \) তাহলে, \( \cos(\frac{\pi}{16}) = \sqrt{\frac{1+\cos(\pi/8)}{2}} \) আমরা আগেই বের করেছি, \( \cos(\frac{\pi}{8}) = \frac{\sqrt{2+\sqrt{2}}}{2} \) তাহলে, \( \cos(\frac{\pi}{16}) = \sqrt{\frac{1+\frac{\sqrt{2+\sqrt{2}}}{2}}{2}} = \sqrt{\frac{2+\sqrt{2+\sqrt{2}}}{4}} = \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \) অতএব, \( 4\cos^2(\frac{\pi}{16}) = 4\left(\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\right)^2 = 4\frac{2+\sqrt{2+\sqrt{2}}}{4} = 2+\sqrt{2+\sqrt{2}} \) সুতরাং, \(4\cos^2(\pi/16) = 2 + \sqrt{2 + \sqrt{2}}\). প্রদত্ত উত্তরটি ভুল। সঠিক উত্তর: \(2 + \sqrt{2 + \sqrt{2}}\) 😅