Another Explanation (5): ```html
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। 🤔🤔🤔
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