已知fx=cos(2x-π/3)+2sin(x-π/4)xsin(x+π/4) 1求fx的最小正周期及单调递增区间;
问题描述:
已知fx=cos(2x-π/3)+2sin(x-π/4)xsin(x+π/4) 1求fx的最小正周期及单调递增区间;
2:求函数fx在区间[-π/12,π/2]上的值域:
答
f(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=2sin(π/6)sin(2x-π/6)=sin(2x-π/6)1.最小正周期T=2π/2=π单增区间2x-π/6∈[2kπ-π/2,2kπ+π/2]x∈[kπ-π/6,kπ+...