已知函数f(x)=cos(2x-pai/3)+2sin(x-pai/4).sin(x+pai/4)求函数在区间[-pai
2个回答

f(x)=cos(2x-π/3)+2sin(x-π/4)cos[π/2-(x+π/4)]

=cos(2x-π/3)+2sin(x-π/4)cos(π/4-x)

=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4)

=cos(2x-π/3)+sin(2x-π/2)

=cos(2x-π/3)-cos2x

=(1/2)cos2x+(√3/2)sin2x-cos2x

=(√3/2)sin2x-(1/2)cos2x

=sin(2x-π/6)

﹣π/12≤x≤π/12

﹣π/6≤2x≤π/6

﹣π/3≤2x-π/6≤0

当2x-π/6=0时,y最大

y(max)=0

当2x-π/6= -π/3时,y最小

y(min)= - √3/2