trigonometry

2011-05-27 6:18 am
If A+B+C=180°,prove that-sin2A+sin2B+sin2C=4sinAcosBcosC

回答 (1)

2011-05-27 6:45 am
✔ 最佳答案
- sin2A + sin2B + sin2C= - 2 sinA cosA + 2 sin(B+C) cos(B-C)= - 2 sinA cosA + 2 sin(180° - A) cos(B-C)= - 2 sinA cosA + 2 sinA cos(B-C)= 2 sinA (cos(B-C) - cosA)= 2 sinA [- 2 sin((B-C + A)/2) sin((B-C - A)/2)]= 2 sinA [- 2 sin((B-C + A + C - C)/2) sin((B-C - A - B + B)/2)]= 2 sinA [- 2 sin((180° - 2C)/2) sin((2B - 180°)/2)]= 2 sinA [- 2 sin(90° - C) sin((B - 90°)]= 2 sinA [ - 2 cosC (- sin(90° - B)) ]= 2 sinA [ - 2 cosC (- cosB) ]= 4sinA cosB cosC


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