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)的最小正周期和图像的对称轴方程
⑵求函数f(x)在区间[-π/12,π/2]上的值域

Answer 1

2sin(x-π/4）sin(x+π/4)
=2sin(x-π/4)cos[π/2-(x-π/4)]
=2sin(x-π/4) cos(π/4-x)
=2sin(x-π/4) cos(x-π/4)
=sin(2x-π/2)
=-sin(π/2-2x)
=-cos2x
∴ f(x)=cos2xcosπ/3+sin2xsinπ/3-cos2x
=-(cos2xcosπ/3-sin2xsinπ/3)
=-cos(2x+π/3)
T=2π/2=π
对称轴2x+π/3=kπ
x=（kπ-π/3）/2
做到这一步，相信下面的你会做了，不懂再问我。

建议楼主多给点分，还有你的三角函数要好好的学习，高考肯定是免不了的

Answer 2

f(x)=cos(2x-π/3）+2sin(x-π/4）sin(x+π/4)
=cos2xcosπ/3+sin2xsinπ/3+2(sinxcosπ/4-cosxsinπ/4)(sinxcosπ/4+cosxsinπ/4)
可以这样打开，不用记太多公式，但计算复杂。

Answer 3

2sin(x-π/4）sin(x+π/4)
=2sin(x-π/4)cos[π/2-(x-π/4)]
=2sin(x-π/4) cos(π/4-x)
=2sin(x-π/4) cos(x-π/4)
=sin(2x-π/2)
=-sin(π/2-2x)
=-cos2x
∴ f(x)=cos2xcosπ/3+sin2xsinπ/3-cos2x
=-(cos2xcosπ/3-sin2xsinπ/3)
=-cos(2x+π/3)
T=2π/2=π