Question

已知函数f(x)=2√3sinxcosx-3sin²x-cos²x+2

若三角形ABC的内角A B C 的对边分别为a b c 且满足b/a=√3，[sin(2A+C)]/sinA=2+2cos(A+C),求f（B）的值

Answer 1

f(x)=2√3sinxcosx-3sin²x-cos²x+2
=√3(2sinxcosx)-(sin²x+cos²x)-(2sin²x+1) +3
=√3sin(2x)-1-cos(2x)+3
=√3sin(2x)-cos(2x) +2
=2[(√3/2)sin(2x)-(1/2)cos(2x)]+2
=2sin(2x-π/6)+2
b/a=√3，由正弦定理得sinB/sinA=√3
sinB=√3sinA
sin(A+C)=sinB=√3sinA
sin(2A+C)/sinA=2+2cos(A+C)
sin[(A+C)+A]/sinA=2-2cosB
[sin(A+C)cosA+cos(A+C)sinA]/sinA=2-2cosB
(√3sinAcosA-cosBsinA)/sinA=2-2cosB
√3cosA-cosB=2-2cosB
cosB=2-√3cosA
sin²B+cos²B=1
(√3sinA)²+(2-√3cosA)²=1
3sin²A+4-4√3cosA+3cos²A=1
2√3cosA=3
cosA=√3/2
cosB=2-√3cosA=2-√3(√3/2)=2-3/2=1/2
B为三角形内角，B=π/3
f(B)=2sin(2·π/3 -π/6)+2
=2sin(π/2)+2
=2·1+2
=4

Answer 2

f(B)=1 B=60, A=30 C=90