2Sin(π/16)=?

প্রশ্ন: \(2\sin(\frac{\pi}{16}) = ?\)

উত্তর: \(\sqrt{2-\sqrt{2+\sqrt{2}}}\)

ব্যাখ্যা:

আমরা জানি, \( \cos(2\theta) = 1 - 2\sin^2(\theta) \).

সুতরাং, \( \sin(\theta) = \sqrt{\frac{1 - \cos(2\theta)}{2}} \).

এখন, \( \theta = \frac{\pi}{16} \) হলে, \( 2\theta = \frac{\pi}{8} \).

সুতরাং, \( \sin(\frac{\pi}{16}) = \sqrt{\frac{1 - \cos(\frac{\pi}{8})}{2}} \).

আবার, \( \cos(\frac{\pi}{8}) = \sqrt{\frac{1 + \cos(\frac{\pi}{4})}{2}} = \sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}} = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} = \frac{\sqrt{2+\sqrt{2}}}{2}\).

অতএব, \( \sin(\frac{\pi}{16}) = \sqrt{\frac{1 - \frac{\sqrt{2+\sqrt{2}}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{2+\sqrt{2}}}{4}} = \frac{\sqrt{2 - \sqrt{2+\sqrt{2}}}}{2} \).

সুতরাং, \( 2\sin(\frac{\pi}{16}) = 2 \cdot \frac{\sqrt{2 - \sqrt{2+\sqrt{2}}}}{2} = \sqrt{2 - \sqrt{2+\sqrt{2}}} \). 🎉

সুতরাং, \(2\sin(\frac{\pi}{16}) = \sqrt{2-\sqrt{2+\sqrt{2}}}\). 🥳