Question

三角函数问题

在三角形abc中，已知sinC／2sinA-sinC=b^2-a^2-c^2／c^2-a^2-b^2
（1）求角B大小
（2）设T=sin^2A+sin^2B+sin^2C求Ｔ的取值范围

Answer 1

1）sinC／2sinA-sinC=b^2-a^2-c^2／c^2-a^2-b^2
sinC=c*sinA/a代入原式得c/（2a-c）=b^2-a^2-c^2／c^2-a^2-b^2
化简得2a(b^2-a^2-c^2+ac)=0,a>0,b^2-a^2-c^2+ac=0
b^2=a^2+c^2-ac;又b^2=a^2+c^2-2accosB
故COSB=1/2,B=60度
2)T=sin^2A+3/4+sin^2(A+B)=sin^2A+3/4+(sinAcos60+cosAsin60)^2
=1+1/4(√3sin2A-cos2A)
=1+1/2sin(2A-φ)(其中tanφ=√3/3)
=1+1/2sin(2A-π/6)
因B=60，那么0<A<2π/3,
则-π/6<2A-π/6<7π/6(显然取π/2时最大，取7π/6或-π/6时取最小值)
故T（3/4，3/2］

Answer 2

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

Answer 3

1）sinC／2sinA-sinC=b^2-a^2-c^2／c^2-a^2-b^2

sinC=c*sinA/a代入原式得c/（2a-c）=b^2-a^2-c^2／c^2-a^2-b^2
化简得2a(b^2-a^2-c^2+ac)=0,a>0,b^2-a^2-c^2+ac=0
b^2=a^2+c^2-ac;又b^2=a^2+c^2-2accosB
故COSB=1/2,B=60度

Answer 4

同学你sinC／2sinA-sinC
是sinC／（2sinA-sinC）还是sinC／2sinA-sinC