Với A, B, C là 3 góc trong 1 tam giác, chứng minh sin A + sin B + sin C = 4 cos A/2
Câu hỏi:
Với A, B, C là 3 góc trong 1 tam giác, chứng minh
sin A + sin B + sin C = \(4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\).
Trả lời:
sin A + sin B + sin C
= \[2\sin \frac{{A + B}}{2}\cos \frac{{A - B}}{2} + 2\sin \frac{C}{2}\cos \frac{C}{2}\]
= \[2\sin \frac{{\pi - C}}{2}\cos \frac{{A - B}}{2} + 2\cos \frac{{\pi - C}}{2}\cos \frac{C}{2}\]
= \[2\cos \frac{C}{2}\cos \frac{{A - B}}{2} + 2\cos \frac{{A + B}}{2}\cos \frac{C}{2}\]
= \[2\cos \frac{C}{2}\left( {\cos \frac{{A - B}}{2} + \cos \frac{{A + B}}{2}} \right)\]
= \[2\cos \frac{C}{2}\left( {2\cos \frac{{A - B + A + B}}{4}\cos \frac{{A - B - \left( {A + B} \right)}}{4}} \right)\]
= \[4\cos \frac{C}{2}\cos \frac{A}{2}.\cos \frac{{ - B}}{2}\]
= \(4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\).
