Question

已知向量a=(cosx+根号3sinx,根号3sinx)b=(cosx-根号3sinx,2cosx)函数fx=ab-cos2x

1）求函数fx的值域
2）若f（c）=1/5，c∈（π/6，π/3）,求sin2c的值

Answer 1

fx=ab-cos2x
=cos²x-3sin²x+2√3sinxcosx-cos2x
=cos²x-sin²x-cos2x+√3sin2x+1-2sin²x-1
=cos2x-cos2x+√3sin2x+cos2x-1
=√3sin2x+cos2x-1
=2(√3/2*sin2x+1/2*cos2x)-1
=2sin(2x+π/6)-1
最大值=2-1=1
最小值=-2-1=-3
∴值域是[-3,1]
(2)
f(C)=2sin(2C+π/6)-1=1/5
sin(2C+π/6)=3/5
∵C∈(π/6,π/3)
∴2C∈(π/3,2π/3)
2C+π/6∈(π/2,5π/6)
∴cos(2C+π/6)<0
cos(2C+π/6)=-4/5
sin2C
=sin(2C+π/6-π/6)
=sin(2C+π/6)*cosπ/6-cos(2C+π/6)*sinπ/6
=3/5*√3/2-(-4/5)*1/2
=(3√3+4)/10
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