2 sin²15° এর মান কত?

Another Explanation (5): প্রথমে, আমরা জানি যে: \[ \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4} \] তাহলে, \[ \sin^2 15^\circ = \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right)^2 = \frac{(\sqrt{6} - \sqrt{2})^2}{16} \] এখন, \[ (\sqrt{6} - \sqrt{2})^2 = (\sqrt{6})^2 - 2 \cdot \sqrt{6} \cdot \sqrt{2} + (\sqrt{2})^2 = 6 - 2 \sqrt{12} + 2 \] কারণ, \[ 2 \sqrt{12} = 2 \times \sqrt{4 \times 3} = 2 \times 2 \sqrt{3} = 4 \sqrt{3} \] তাই, \[ (\sqrt{6} - \sqrt{2})^2 = 6 + 2 - 4 \sqrt{3} = 8 - 4 \sqrt{3} \] অতএব, \[ \sin^2 15^\circ = \frac{8 - 4 \sqrt{3}}{16} = \frac{8}{16} - \frac{4 \sqrt{3}}{16} = \frac{1}{2} - \frac{\sqrt{3}}{4} \] এখন, \[ 2 \sin^2 15^\circ = 2 \times \left( \frac{1}{2} - \frac{\sqrt{3}}{4} \right) = 1 - \frac{\sqrt{3}}{2} \] অর্থাৎ, \[ 2 \sin^2 15^\circ = 1 - \frac{\sqrt{3}}{2} \] অথবা, \[ 2 \sin^2 15^\circ = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \] সুতরাং, **উত্তর** হলো: ```html \frac{2 - \sqrt{3}}{2} ```