求函数f(x)=2cos(x+π/4)cos(x-π/4)+更号3sinx的值域.

f(x)=2cos(x+π/4)cos(x-π/4)+√3sinx
=2cos(x+π/2-π/4)cos(x-π/4)+√3sinx
=2cos(π/2-π/4+x)cos(x-π/4)+√3sinx
=2cos[π/2-(π/4-x)]cos(x-π/4)+√3sinx
=2sin(π/4-x)cos(x-π/4)+√3sinx
=2sin(π/4-x)cos(π/4-x)+√3sinx
=sin(π/2-2x)+√3sinx
=cos2x+√3sinx
=1-2sin²x+√3sinx
=-2sin²x+√3sinx+1
=-2(sin²x-√3/2*sinx)+1
=-2(sin²x-√3/2*sinx+3/16)+1+3/8
=-2(sinx-√3/4)²+11/8
f(x)的值域:(√3-1,11/8)

f(x)=2cos(x+π/4)cos(x-π/4)+更号3sinx=cos(2x)+cosπ/2+√3sinx=1-2sin²x+√3sinx=-2(sin²x-√3/2sinx)+1=-2(sinx-√3/4)²+11/8所以当sinx=√3/4时,取最大值=11/8sinx=-1时取最小值=1-2-√3=-1-√...