求函数y=1/2cos^2x+根号3/2 sinxcosx+1最大值及取最大值时X的值集合

1个回答

  • 1/2=sin(π/6),√3/2=cos(π/6),因此可对表达式化简:

    y=(1/2)(cosx)^2+(√3/2)sinxcosx+1

    =cosx[sin(π/6)cosx+cos(π/6)sinx]+1

    =sin(x+π/6)cosx+1 ………………………………………………………(1)

    sin(2x+π/6)=sin(x+π/6+x)=sin(x+π/6)cosx+cos(x+π/6)sinx ………(2)

    1/2=sin(π/6)=sin(x+π/6-x)=sin(x+π/6)cosx-cos(x+π/6)sinx ………(3)

    (2)+(3)可得:sin(x+π/6)cosx=[sin(2x+π/6)]/2+1/4 ……………(4)

    把(4)代入(1)继续化简:

    sin(x+π/6)cosx+1

    =[sin(2x+π/6)]/2+1/4+1

    =[sin(2x+π/6)]/2+5/4

    因此:y=[sin(2x+π/6)]/2+5/4

    y取最大值时,sin(2x+π/6)=1,即2x+π/6=2kπ+π/2,求得x=kπ+π/6(k∈Z),

    因此所求x的集合为:{x|x=kπ+π/6(k∈Z)}