যদি A + B + C = pi  হয়, তবে sinA + sinB + sinC = ?

Explanation:

Another Explanation (5): যদি \( A + B + C = \pi \) হয়, তবে \( \sin A + \sin B + \sin C = ? \) আমরা জানি, \( \sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \) তাহলে, \(\sin A + \sin B + \sin C = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} + \sin C \) যেহেতু \( A + B + C = \pi \), তাই \( A + B = \pi - C \) সুতরাং, \( \frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2} \) এখন, \(\sin A + \sin B + \sin C = 2 \sin (\frac{\pi}{2} - \frac{C}{2}) \cos \frac{A-B}{2} + \sin C \) \( = 2 \cos \frac{C}{2} \cos \frac{A-B}{2} + 2 \sin \frac{C}{2} \cos \frac{C}{2} \) \( = 2 \cos \frac{C}{2} (\cos \frac{A-B}{2} + \sin \frac{C}{2}) \) \( = 2 \cos \frac{C}{2} (\cos \frac{A-B}{2} + \sin (\frac{\pi}{2} - \frac{A+B}{2})) \) \( = 2 \cos \frac{C}{2} (\cos \frac{A-B}{2} + \cos \frac{A+B}{2}) \) আমরা জানি, \( \cos(x) + \cos(y) = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} \) সুতরাং, \( \cos \frac{A-B}{2} + \cos \frac{A+B}{2} = 2 \cos \frac{\frac{A-B}{2} + \frac{A+B}{2}}{2} \cos \frac{\frac{A-B}{2} - \frac{A+B}{2}}{2} \) \( = 2 \cos \frac{A}{2} \cos \frac{-B}{2} = 2 \cos \frac{A}{2} \cos \frac{B}{2} \) তাহলে, \(\sin A + \sin B + \sin C = 2 \cos \frac{C}{2} (2 \cos \frac{A}{2} \cos \frac{B}{2}) \) \( = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \) অতএব, \( \sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \) 🎉