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Composite Function

If A + B + C ...

Question

If $A + B + C = π$ , then prove that $s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - 1 = 4 s i n \frac{π - 4}{4} . s i n \frac{π - B}{4} . s i n \frac{π - C}{4}$

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Solution

$A + B + C = π \Rightarrow A + B = π - C (1) B + C = π - A (2) A + C = π - B (3)$
To Prove: $s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - 1 = 4 s i n \frac{π - 4}{4} . s i n \frac{π - B}{4} . s i n \frac{π - C}{4} L H S = s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - 1 = s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - s i n \frac{π}{2} (∵ 1 = s i n \frac{π}{2}) = 2 s i n \frac{\frac{A}{2} + \frac{B}{2}}{2} c o s \frac{\frac{A}{2} - \frac{B}{2}}{2} + 2 c o s \frac{\frac{C}{2} + \frac{π}{2}}{2} s i n \frac{\frac{C}{2} - \frac{π}{2}}{2} = 2 s i n \frac{A + B}{4} c o s \frac{A - B}{4} + 2 c o s \frac{C + π}{4} s i n \frac{C - π}{4} = 2 s i n \frac{π - C}{4} c o s \frac{A - B}{4} + 2 c o s \frac{C + π}{4} (- s i n \frac{π - C}{4}) = 2 s i n \frac{π - C}{4} [c o s \frac{A - B}{4} - c o s \frac{C + π}{4}] = 2 s i n \frac{π - C}{4} [- 2 s i n \frac{\frac{A - B}{4} + \frac{C + π}{4}}{2} s i n \frac{A - B}{4} - \frac{C + π}{2}] = - 4 s i n \frac{π - C}{4} [s i n \frac{(A + C) - B + π}{8} s i n \frac{A + π - (B + C)}{8}] = - 4 s i n \frac{π - C}{4} [s i n \frac{(A + C) - B + π}{8} s i n \frac{A + π - (π - 4)}{8}]$
$= - 4 s i n \frac{π - C}{4} [s i n \frac{2 π - 2 B}{8} s i n \frac{2 A - 2 π}{8}] = - 4 s i n \frac{π - C}{4} [s i n \frac{π - B}{4} s i n \frac{A - π}{4}] = 4 s i n \frac{π - C}{4} s i n \frac{π - B}{4} 4 s i n \frac{π - A}{4} = R H S$
Hence Proved.

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