Question

已知函数f(x)=2(cosx)^2+2√3sinxcosx,（1）求函数f(x)在[-π/6,π/3]上的值域

（2）在△ABC中，若f(C)=2,2sinB=cos(A-C)-cos(A+C),求tanA的值

Answer 1

1.
f(x)=2(cosx)^2+2√3sinxcosx
=cos2x+1+√3sin2x
=2sin(2x+π/6)+1
=2sin(2(x+π/12))+1
[-π/6,π/3]上,
f(-π/6)=2*(-1/2)+1=0
f(π/3)=2*sin(5π/6)+1=2
f(x)最大=f(π/6)=2+1=3
f(x)在[-π/6,π/3]上的值域:[0,3]

2.
f(C)=2
2sin(2(C+π/12))+1=2
sin(2(C+π/12))=1/2
2(C+π/12)=π/6, 或2(C+π/12)=5π/6
C=0（舍弃），或C=π/3

2sinB=cos(A-C)-cos(A+C)
2sinB=-2sinAsin(-C)
sinB=sinAsinC
sin(A+C)=√3/2 sinA
sinA*1/2+cosA)*√3/2=√3/2 sinA
tanA=sinA/cosA=√3/(√3-1)=(3+√3)/2