Question

设函数f(x)=cos(2x+π/3)+sin^2x，求函数的最大值和最小正周期

Answer 1

f(x)=cos(2x+π/3)+sin^2x
=cos2xcosπ/3-sin2xsinπ/3+sin^2x
=1/2cos2x+(1-√3/2)sin2x
=√[(1/2)^2+(1-√3/2)^2]sin(2x+Φ)
=(4-2√3)/2sin(2x+Φ)
其中，sinΦ=1/(4-2√3);cosΦ=(2-√3)/(4-2√3);
所以，f(x)(max) =(4-2√3)/2;
T=2π/2=π.

Answer 2

f(x)=(1/2)cos2x-(√3/2)sin2x+(1/2)(1-cos2x)=-(√3/2)sin2x+1/2，
最大值为(1+√3)/2，最小正周期为π。