已知函数f(x)=√3sin(wx)-2sin^2(wx/2)+m(w>0)的最小正周期为3π

题目

已知函数f(x)=√3sin(wx)-2sin^2(wx/2)+m(w>0)的最小正周期为3π
且当x∈【0,π】时,函数f(x)的最小值为0.
(1)求函数f(x)的表达式
（2）在三角形ABC,若f（C）=1,且2sin^2 B=cosB+cos(A-C),求sinA的值

答案

(1) f(x)=√3sin(wx)+1-2sin^2(wx/2)+m-1=√3sin(wx)+cos(wx)+m-1
=2sin(wx+π/6)+m-1
∵T=2π/w=3π==>w=2/3
∴f(x)=2sin(2/3x+π/6)+m-1
令F’(x)=4/3cos(2/3x+π/6)=0==>2/3x+π/6=2kπ-π/2==>x=3kπ-π；
2/3x+π/6=2kπ+π/2==>x=3kπ+π/2；
F”(x)=-8/9sin(2/3x+π/6)
F”(-π)=-8/9sin(-2π/3+π/6)=8/9>0
∴f(x)在x=3kπ-π取极小值,在x=3kπ+π/2取极大值；
又当x∈【0,π】时,函数f(x)的最小值为0
f(0)=2sin(2/3x+π/6)+m-1=m,f(π)=2sin(2/3x+π/6)+m-1=m
∴m=0
∴f(x)=2sin(2/3x+π/6)-1
(2)∵在三角形ABC,若f（C）=1
f(C)=2sin(2/3C+π/6)-1=1==>2/3C+π/6=π/2==>C=π/2
又2(sinB)^2=cosB+cos(A-C)
2(sin(π/2-A))^2=cos(π/2-A)+cos(A-π/2)
2(cosA)^2=sinA+sinA==>(cosA)^2=sinA==>1-(sinA)^2=sinA
(sinA)^2+sinA-1=0
∴sinA=(√5-1)/2