已知函数 y=sin(2x-(π/3))+cos(2x-(π/6))+2cos²x-1,x∈R

问题描述:

已知函数 y=sin(2x-(π/3))+cos(2x-(π/6))+2cos²x-1,x∈R
1.求函数f(x)的最小正周期!
2.求函数f(x)在区间[-π/4,π/4]上的最大值和最小值.

y=sin(2x-(π/3))+cos(2x-(π/6))+2cos²x-1;
=1/2sin2x-√3/2cos2x+√3/2cos2x+1/2sin2x+2cos²x-1;
=sin2x+2cos²x-1;
=sin2x+cos2x;
=√2sin(2x+π/4);
1.最小正周期=2π/2=π;
2.x∈[-π/4,π/4];
2x+π/4∈[-π/4,3π/4];
当2x+π/4=-π/4时;
f(x)有最小值=√2*(-√2/2)=-1;
当2x+π/4=π/2时;
f(x)有最大值=√2*1=√2;
如果本题有什么不明白可以追问,