যদি  A+B+C= π  হয় তবে  sin^2(A/2) + sin^2(B/2)+ sin^2(C/2) =? 

দেওয়া আছে, \(A + B + C = \pi\)

আমরা জানি, \(sin^2(x) = \frac{1 - cos(2x)}{2}\)

সুতরাং,

\(sin^2(A/2) + sin^2(B/2) + sin^2(C/2) = \frac{1 - cosA}{2} + \frac{1 - cosB}{2} + \frac{1 - cosC}{2}\)

\(= \frac{3}{2} - \frac{1}{2} (cosA + cosB + cosC)\)

এখন, \(cosA + cosB + cosC = cosA + cosB + cos(\pi - (A+B))\)

\(= cosA + cosB - cos(A+B)\)

\(= 2cos(\frac{A+B}{2})cos(\frac{A-B}{2}) - [2cos^2(\frac{A+B}{2}) - 1]\)

\(= 2cos(\frac{A+B}{2})cos(\frac{A-B}{2}) - 2cos^2(\frac{A+B}{2}) + 1\)

যেহেতু \(A+B+C = \pi\), তাই \(\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}\)

\(cos(\frac{A+B}{2}) = cos(\frac{\pi}{2} - \frac{C}{2}) = sin(\frac{C}{2})\)

সুতরাং, \(cosA + cosB + cosC = 2sin(\frac{C}{2})cos(\frac{A-B}{2}) - 2sin^2(\frac{C}{2}) + 1\)

\(= 1 + 2sin(\frac{C}{2})[cos(\frac{A-B}{2}) - sin(\frac{C}{2})]\)

\(= 1 + 2sin(\frac{C}{2})[cos(\frac{A-B}{2}) - sin(\frac{\pi}{2} - \frac{A+B}{2})]\)

\(= 1 + 2sin(\frac{C}{2})[cos(\frac{A-B}{2}) - cos(\frac{A+B}{2})]\)

\(= 1 + 2sin(\frac{C}{2})[2sin(\frac{A}{2})sin(\frac{B}{2})]\)

\(= 1 + 4sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})\)

অতএব,

\(sin^2(A/2) + sin^2(B/2) + sin^2(C/2) = \frac{3}{2} - \frac{1}{2} [1 + 4sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})]\)

\(= \frac{3}{2} - \frac{1}{2} - 2sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})\)

\(= 1 - 2sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})\) 🎉