Question

已知函数f（x）= -根号3sin²x+sinxcosx

（1）球f（π/6）的值 （2）求函数f（x）的最小正周期及最大值

Answer 1

f（π/6）=-√3sin²(π/6)+sin(π/6)cos(π/6)
=-√3*(1/4)+1/2*(√3/2)
=0

f(x)=-√3/2(1-cos2x)+1/2sin2x
=1/2sin2x+√3/2cos2x-√3/2
=sin(2x+π/3)-√3/2
T=2π/2=π
当sin(2x+π/3)=1的时候
f(x)有最大值
此时f(x)的最大值=1-√3/2=（2-√3）/2

Answer 2

1、
f(x)=-√3/2*(1-cos2x)+1/2*sin2x
=1/2*sin2x+√3/2*cos2x-√3/2
=sin2xcosπ/3+cos2xsinπ/3-√3/2
=sin(2x+π/3)-√3/2
所以f(π/6)=sin2π/3-√3/2=0

2、
f(x)=sin(2x+π/3)-√3/2
所以T=2π/2=π
最大值=1-√3/2

Answer 3

f(x)= -√3/2(1-cos2x)+1/2sin2x
=-√3/2sin(2x+∏/3)
f（π/6）=-√3/2(√3/2)=-3/2
T=2∏/2=∏
f(x)max=√3/2