যদি cosα + cosβ = a, sinα + sinβ = b এবং α + β = 2θ হয় তবে cos3θ/cosθ এর মান কত?

দেওয়া আছে: \( cos\alpha + cos\beta = a \) \( sin\alpha + sin\beta = b \) \( \alpha + \beta = 2\theta \) এখন, আমরা জানি, \( cosC + cosD = 2cos(\frac{C+D}{2})cos(\frac{C-D}{2}) \) \( sinC + sinD = 2sin(\frac{C+D}{2})cos(\frac{C-D}{2}) \) সুতরাং, \( a = cos\alpha + cos\beta = 2cos(\frac{\alpha+\beta}{2})cos(\frac{\alpha-\beta}{2}) = 2cos\theta cos(\frac{\alpha-\beta}{2}) \) \( b = sin\alpha + sin\beta = 2sin(\frac{\alpha+\beta}{2})cos(\frac{\alpha-\beta}{2}) = 2sin\theta cos(\frac{\alpha-\beta}{2}) \) এখন, \( a^2 + b^2 = (2cos\theta cos(\frac{\alpha-\beta}{2}))^2 + (2sin\theta cos(\frac{\alpha-\beta}{2}))^2 \) \( = 4cos^2\theta cos^2(\frac{\alpha-\beta}{2}) + 4sin^2\theta cos^2(\frac{\alpha-\beta}{2}) \) \( = 4cos^2(\frac{\alpha-\beta}{2})(cos^2\theta + sin^2\theta) \) \( = 4cos^2(\frac{\alpha-\beta}{2}) \) অতএব, \( cos^2(\frac{\alpha-\beta}{2}) = \frac{a^2+b^2}{4} \) এখন, \( \frac{a}{b} = \frac{2cos\theta cos(\frac{\alpha-\beta}{2})}{2sin\theta cos(\frac{\alpha-\beta}{2})} = \frac{cos\theta}{sin\theta} = cot\theta \) আমাদের নির্ণয় করতে হবে \( \frac{cos3\theta}{cos\theta} \) এর মান। আমরা জানি, \( cos3\theta = 4cos^3\theta - 3cos\theta \) সুতরাং, \( \frac{cos3\theta}{cos\theta} = \frac{4cos^3\theta - 3cos\theta}{cos\theta} = 4cos^2\theta - 3 \) এখন, \( cos^2(\frac{\alpha-\beta}{2}) = \frac{a^2+b^2}{4} \) \( cos(\frac{\alpha-\beta}{2}) = \pm\frac{\sqrt{a^2+b^2}}{2} \) যেহেতু, \( a = 2cos\theta cos(\frac{\alpha-\beta}{2}) \) \( cos\theta = \frac{a}{2cos(\frac{\alpha-\beta}{2})} = \frac{a}{2(\pm\frac{\sqrt{a^2+b^2}}{2})} = \pm\frac{a}{\sqrt{a^2+b^2}} \) সুতরাং, \( cos^2\theta = \frac{a^2}{a^2+b^2} \) তাহলে, \( \frac{cos3\theta}{cos\theta} = 4cos^2\theta - 3 = 4\frac{a^2}{a^2+b^2} - 3 = \frac{4a^2 - 3a^2 - 3b^2}{a^2+b^2} = \frac{a^2 - 3b^2}{a^2+b^2} \) 🤔🤔🤔 পূর্বের উত্তরের সাথে মিলছে না। অন্যভাবে চেষ্টা করি। \( a^2 + b^2 = 4cos^2(\frac{\alpha-\beta}{2}) \) থেকে আমরা লিখতে পারি, \( cos^2(\frac{\alpha-\beta}{2}) = \frac{a^2+b^2}{4} \) \( cos(\alpha - \beta) = 2cos^2(\frac{\alpha-\beta}{2}) - 1 = 2(\frac{a^2+b^2}{4}) - 1 = \frac{a^2+b^2}{2} - 1 \) এখন, \( cos3\theta = 4cos^3\theta - 3cos\theta \) সুতরাং, \( \frac{cos3\theta}{cos\theta} = 4cos^2\theta - 3 \) আমরা জানি, \( a^2 + b^2 = (cos\alpha + cos\beta)^2 + (sin\alpha + sin\beta)^2 \) \( = cos^2\alpha + 2cos\alpha cos\beta + cos^2\beta + sin^2\alpha + 2sin\alpha sin\beta + sin^2\beta \) \( = (cos^2\alpha + sin^2\alpha) + (cos^2\beta + sin^2\beta) + 2(cos\alpha cos\beta + sin\alpha sin\beta) \) \( = 1 + 1 + 2cos(\alpha - \beta) = 2 + 2cos(\alpha - \beta) \) \( = 2 + 2cos(\alpha - \beta) \) সুতরাং, \( cos(\alpha - \beta) = \frac{a^2+b^2}{2} - 1 \) আবার, \( cos2\theta = cos(\alpha + \beta) = cos\alpha cos\beta - sin\alpha sin\beta \) \( a^2+b^2 = 2 + 2(cos\alpha cos\beta + sin\alpha sin\beta) \) \( \frac{a^2+b^2-2}{2} = cos(\alpha - \beta) \) কিন্তু আমাদের দরকার \( 4cos^2\theta - 3 \) এর মান। যদি \( cos^2(\frac{\alpha-\beta}{2}) = \frac{a^2+b^2}{4} \) হয়, তবে \( cos\theta = \frac{a}{2cos(\frac{\alpha-\beta}{2})} \) এই মুহূর্তে আমার মনে হচ্ছে প্রশ্নটিতে অথবা উত্তরে কোথাও ভুল আছে। 🤔🤔🤔 ```