Question

已知函数f（x）sinxcosx-√3sin²x+√3/2（1）求函数f（x）的最小正周期

在f（x）在[-π/2，π/2]上的单调减区间；（3）求f（x）在[0，π/4]上的最值及相应的x取值范围

Answer 1

答：
f(x)=sinxcosx-√3sinx*sinx+√3/2
=(1/2)sin2x+(√3/2)cos2x
=sin(2x+π/3)
（1）f(x)的周期为kπ，所以最小正周期为π

（2）-π/2<=x<=π/2

-2π/3<=2x+π/3<=4π/3
所以f(x)在[-π/2,π/2]上的单调减区间为：-2π/3<=2x+π/3<=-π/2及 π/2<=2x+π/3<=4π/3
即单调减区间为[-π/2,-5π/12]及[π/12,π/2]

（3）0<=x<=π/4，π/3<=2x+π/3<=5π/6
f(x)min=sin(5π/6)=1/2，此时x=π/4；
f(x)max=sin(π/2)=1，此时x=π/12.
.

Answer 2

f(x)=sinxcosx-√3sin²x+√3/2=(1/2)sin2x - (1/2)(1-cos2x)+√3/2
=（1/2）（sin2x+cos2x-1+√3)
=(1/2)(√2sin(2x+π/4)-1+√3)

这样的话最小正周期是π
单调减区间和最值求导易求

难度只在化简上 要灵活运用二倍角公式

追问

在f（x）在[-π/2，π/2]上的单调减区间；（3）求f（x）在[0，π/4]上的最值及相应的x取值范围