函数y=sinxsin(π/3 - x)的值域为？？

这种题用的都是积化和差
y=sinxsin(π/3 - x)=-1/2[cos(π/3)-cos(2x-π/3)]=-1/4+1/2cos(2x-π/3)
到这里值域很清楚了。
-1≤cos(2x-π/3)≤1，
-3/4≤-1/4+1/2cos(2x-π/3)≤1/4

y=sinx(√3/2*cosx-1/2*sinx)
=√3/2*sinxcosx-1/2*sin²x
=(√3/4)sin2x-(1/2)*(1-cos2x)/2
=(√3/4)sin2x+(1/4)cosx-1/2
=√[(√3/4)²+(1/4)²]sin(2x+z)-1/2
=(1/2)sin(2x+z)-1/2
其中tanz=(1/4)/(√3/4)=1/√3

-1<=sin(2x+z)<=1
所以-1/2<=(1/2)sin(2x+z)<=1/2
-1/2-1/2<=(1/2)sin(2x+z)-1/2<=1/2-1/2
所以值域[-1,0]

-sin1到sin1