Question

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

Answer 1

f(x)=cos(2x+π/3)+sin²x
=cos 2xcos π/3-sin 2xsin π/3 +sin²x
=1/2cos 2x-√3/2sin 2x +sin²x
=1/2cos 2x-√3/2sin 2x + -（-2sin²+1-1）/2
=1/2cos 2x-√3/2sin 2x -（cos 2x-1)/2
=1/2cos 2x-√3/2sin 2x -1/2cos 2x+1/2
=-√3/2sin 2x +1/2

当sin 2x取最小值-1时，f（x）取最大值（√3+1）/2

最小正周期为T=2π/2=π

Answer 2

f(x)=cos2x*1/2-√3*sin2x+(1-cos2x)/2
=cos2x-√3sin2x+1/2
=2cos(2x+π/3)+1/2

最小正周期T=2π/2=π
当cos（2x+π/3）=1
取最大值为5/2

Answer 3

f(x)=cos(2x+π/3)+sin²X
=1/2*cos2x-√3/2*sin2x+(1/2)(1-cos2x)
=1/2-√3/2*sin2x,
f(x)的最大值=（1+√3）/2.
最小正周期=π.