Question

已知函数f（x）=cos（2x+π/3）+sin²x-cos²x+2√3sinxcosx求函数f（x）的单调递减区间

已知函数f（x）=cos（2x+π/3）+sin²x-cos²x+2√3sinxcosx
1.求函数f（x）的单调递减区间
2.若x∈[0，π/2]，求f（x）的最值
3.若f（α）=1/7,2α是第一象限角，求sin2α的值

Answer 1

f(x)=cos(2x+π/3)-cos2x+(√3)sin2x
=cos(2x+π/3)-2[(1/2)cos2x-(√3/2)sin2x]
=cos(2x+π/3)-2[cos2xcos(π/3)-sin2xsin(π/3)]
=cos(2x+π/3)-2cos(2x+π/3)=-cos(2x+π/3)
①.单调区间：
由 2kπ≤2x+π/3≤2kπ+π，得单增区间为：[kπ-π/6，kπ+π/3];
由 2kπ-π≤2x+π/3≤2kπ，得单减区间为：[kπ-2π/3，kπ-π/6];
②.在x∈[0，π/2]内，minf(x)=f(0)=-cos(π/3)=-1/2;
maxf(x)=f(π/3)=-cos(2π/3+π/3)=-cosπ=1.
③. f(α)=-cos(2α+π/3)=1/7
cos(2α+π/3)=-1/7; 2α∈第一象限，∴(2α+π/3)∈第二象限。
2α+π/3=arccos(-1/7)=π-arccos(1/7)
2α=(2π/3)-arccos(1/7)
∴sin2α=sin[(2π/3)-arccos(1/7)]=sin(2π/3)(1/7)-cos(2π/3)√(1-1/49)
=(1/7)sin(π/3)+(1/2)(4√3/7)=(√3)/14+2√3/7=5(√3)/14.