Question

求第18题详细过程

Answer 1

（1）
f(x)=cos(2x-π/3)+2sin(x-π/4)cos(x-π/4) √ π/
=cos(2x-π/3)+sin(2x-π/2)（sin的二倍角公式）
=cos(2x-π/3)+cos2x
=cos2xcosπ/3+sin2xsinπ/3+cos2x
=1/2cos2x+√3/2sin2x+cos2x
=3/2cos2x+√3/2sin2x
=√3(√3/2cos2x+1/2sin2x)
=√3(sin2xcosπ/3+cos2xsinπ/3)
=√3sin(2x+π/3)

所以f(x)的最小正周期为2π/2=π

（2） -π/12≤x≤π/12
-π/6≤2x≤π/6
π/6≤2x+π/3≤π/2
sinx在[π/6,π/2]是递增的，sinπ/6=1/2; sinπ/2=1
所以，f(x)在[-π/12,π/12]的值域为[1/2,1]