Question

在三角形ABC中，角A,B,C,所对边分别为a,b,c,已知sinC/2sinA－sinC＝b平方－a平方－c平方/c平方－a平方...

在三角形ABC中，角A,B,C,所对边分别为a,b,c,已知sinC/2sinA－sinC＝b平方－a平方－c平方/c平方－a平方－b平方。求角B大小。设T＝sin平方A＋sin平方B＋sin平方C，求T的取值范围

Answer 1

sinC/(2sinA-sinC) = (b²-a²-c²)／(c²-a²-b²)
c/(2a-c) = (b²-a²-c²)／(c²-a²-b²)
(2a-c)/c = (c²-a²-b²)／(b²-a²-c²)
(2a-c)/c +1 = (c²-a²-b²)／(b²-a²-c²) +1
2a/c = [(c²-a²-b²)+(b²-a²-c²)]／(b²-a²-c²)
2a/c = -2a² / (b²-a²-c²)
a²+c²-b² = ac
2accosB=ac
cosB=1/2
B= 60°

T = sin²A+sin²B+sin²C , b² = (a²+c²-ac)
= 3 - (cos²A+cos²B+cos²C)
= 3 - [(a²+b²-c²)²/4a²b² + (a²+c²-b²)²/4a²c² + (b²+c²-a²)²/4b²c²]
= 3 - [(a²+a²+c²-ac-c²)²/4a²b² + (a²+c²-a²-c²+ac)²/4a²c² + (a²+c²-ac+c²-a²)²/4b²c²]
= 3 - [(2a²-ac)²/4a²b² + 1/4 + (2c²-ac)²/4b²c²]
= 11/4 - [ (2a-c)²/4b² + (2c-a)²/4b²]
= 11/4 - [(5a²-8ac+5c²)/4(a²+c²-ac)]
= 11/4 - 5/4 + 3ac/4(a²+c²-ac)
= 3/2 + 3ac/4(a²+c²-ac)
≤ 3/2 + 3ac/4(2ac-ac) = 9/4

因此，当a或c近似为0时，T取最小值无限接近2/3
当a=c时，T取最大值9/4 ,即：2/3<T≤9/4