sin 67 1/2^o=? 

Explanation:

Another Explanation (5): sin 67.5° এর মান নির্ণয়: আমরা জানি, \(sin θ = \sqrt{\frac{1 - cos 2θ}{2}}\) এখানে, θ = 67.5° সুতরাং, 2θ = 2 × 67.5° = 135° এখন, \(cos 135° = cos (180° - 45°) = -cos 45° = -\frac{1}{\sqrt{2}}\) অতএব, \(sin 67.5° = \sqrt{\frac{1 - (-\frac{1}{\sqrt{2}})}{2}}\) \(= \sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}}\) \(= \sqrt{\frac{\frac{\sqrt{2} + 1}{\sqrt{2}}}{2}}\) \(= \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}}\) এখন, লব ও হরকে \(\sqrt{2}\) দিয়ে গুণ করে পাই, \(= \sqrt{\frac{\sqrt{2}(\sqrt{2} + 1)}{2\sqrt{2} \cdot \sqrt{2}}}\) \(= \sqrt{\frac{2 + \sqrt{2}}{4}}\) \(= \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{4}}\) \(= \frac{\sqrt{2 + \sqrt{2}}}{2}\) \(= \frac{1}{2}\sqrt{2 + \sqrt{2}}\) সুতরাং, \(sin 67\frac{1}{2}° = \frac{1}{2}\sqrt{2 + \sqrt{2}}\) 🎉🎉🎉