Prove that: $sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = 4 sinAsinBsinC$

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Solution

$\begin{matrix} T P : sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = 4 sinAsinBsinC L e t, B + C = π - A = sin (π - A - A) + sin (π - B - B) + sin (π - C - C) = sin (2 A) + sin (2 B) + sin (2 C) = 2 sin (A + B) c o s (A - B) + 2 sin C . c o s C = 2 sin C . c o s (A - B) + 2 s i n C . c o s C = 2 sin C [c o s (A - B) + 2 c o s C] = 2 sin C [c o s (A - B) + 2 c o s (π - (A + B))] = 2 sin C [c o s (A - B) - c o s (A + B)] = 2 sin C [cosAcosB + s i n A s i n B - c o s A c o s B + s i n A s i n B] = 4 sin A s i n B s i n C \end{matrix}$

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