Question

已知三角形ABC中的内角A,B,C的对边分别为a,b,c,定义向量m=(2sinB,-根号3)，定量n=(cos2B,2cos平方B/2-1)且

定量m//定量n （1）求角B的大小; (2)如果b=2,求三角形ABC的面积的最大值是？

Answer 1

(1)
m//n
=>
2sinB/(-√3 ) = cos2B/(2[cos(B/2)]^2-1)
2sinB(2[cos(B/2)]^2-1) =-√3cos2B
sin2B = -√3cos2B
tan2B = -√3
B = π/3
(2)
b=2
by sine-rule
a/sinA = b/sinB
a= (b/sinB)sinA = (4√3/3)sinA
c= (b/sinB)sinC = (4√3/3)sinC = (4√3/3)sin(2π/3-A)
A1= 三角形ABC的面积
= (1/2)ac sinB
= (1/2)(4√3/3)sinA .(4√3/3)sin(2π/3-A) (√3/2)
= (4/3)sinAsin(2π/3-A)
= (2/3)[cos(2A-2π/3)-cos(2π/3)]
=(2/3)[cos(2A-2π/3)+1/2]
max A1 at A=π/3
=(2/3)[1+1/2]
=1

Answer 2

楼下的错了
2sinB/(2[cos(B/2)]^2-1)= (-√3 ) cos2B
2BcosB +√3cos2B=0
2sin（2B+π/3）=0
因为在三角形中 所以 2B+π/3=0 B=Kπ/2-π/6 B=π/3或π5/6