Question

已知函数f(x)=√3sin(ωx)-2[sin(ωx/2)]^2的最小正周期为3π，

1.当x属于[π/2，3π/4]时，求函数f(x)最小值
2.在三角形ABC中，若f（C）=1，且2sin^B=cos(A-C)+cosB，求sinA的值

Answer 1

f(x)=√3sin(ωx)-2[sin(ωx/2)]^2=√3sin(ωx)-[1-cos(ωx)]
=√3sin(ωx)+cos(ωx)-1=2{[(√3)/2]sin(ωx)+(1/2)cos(ωx)}-1=2sin(ωx+π/6)-1
最小正周期为3π=2π/ω,ω=2/3，
f(x)=2sin(2x/3+π/6)-1
1.当π/2=<x<=3π/4时，
π/3=<2x/3<=π/2,
π/2=π/6+π/3=<2x/3+π/6<=π/2+π/6=2π/3,
(根号3)/2=sin(2π/3)=<sin(2x/3+π/6)<=sin(π/2)=1,
根号3=<2sin(2x/3+π/6)<=2
根号3-1=<2sin(2x/3+π/6)-1<=1
f(x)最小值=根号3-1
2. f(C)=2sin(2C/3+π/6)-1=1，
2sin(2C/3+π/6)=2，
sin(2C/3+π/6)=1，
0<C<π
0<2C/3<2π/3，
π/6<2C/3+π/6<2π/3+π/6=5π/6，
2C/3+π/6=π/2,
C=π/2,
A+B=π/2,
sinB=cosA,cosB=sinA,
由2sinB=cos(A-C)+cosB，
得2cosA=cos(A-π/2)+sinA,
2cosA=sinA+sinA,
cosA=sinA,tanA=1,A=π/4,