Question

已知函数f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)

⑴求函数f(x)的最小正周期
⑵求函数f(x)在区间[-π/12,π/2]上的值域

当中2sin(x-π/4）sin(x+π/4)的具体化简步骤是什么。用到什么公式，怎么代的可以具体说一下吗。

Answer 1

f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)
=cos2xcosπ/3+sin2xsinπ/3+2sin(x-π/4）sin(x+π/2-π/4)
=1/2*cos2x+√3/2*sin2x+2sin(x-π/4）sin[π/2-(π/4-x)]
=1/2*cos2x+√3/2*sin2x+2sin(x-π/4)cos(π/4-x)
=1/2*cos2x+√3/2*sin2x+2sin(x-π/4)cos(x-π/4)(倍角公式)
=1/2*cos2x+√3/2*sin2x+sin[2(x-π/4)]
=1/2*cos2x+√3/2*sin2x+sin(2x-π/2)
=1/2*cos2x+√3/2*sin2x-sin(π/2-2x)
=1/2*cos2x+√3/2*sin2x-cos2x
=√3/2*sin2x-1/2*cos2x
=sin2xcosπ/6-cos2xsinπ/6
=sin(2x-π/6)
T=2π/2=π
x∈[-π/12,π/2]
2x∈[-π/6,π]
2x-π/6∈[-π/3,5π/6]
f(x)∈[-√3/2,1]