Question

求函数y=sin^2x+根号3sinxcosx-1的值域

Answer 1

y=(sinx)^2+(根号3)sinxcosx-1
=(1/2)[2(sinx)^2+2(根号3)sinxcosx-2]
=(1/2)[(根号3)sin2x-cos2x-1]
=(根号3/2)sin2x-(1/2)cos2x-1/2
=cos(派/6)sin2x-sin(派/6)cos2x-1/2
=sin[2x-(派/6)]-1/2
因为 -1<=sin[2x-(派/6)]<=1
所以 -1-1/2<=sin[2x-(派/6)]-1/2<=1-1/2
即： -3/2<=y<=1/2
所以 函数y的值域是[-3/2，1/2]。

Answer 2

y=sin^2x+根号3sinxcosx-1=根号3sinxcosx-（1-2sin^2x）/2-1/2=根号3/2×sin2x-cos2x/2-1/2=sin2xcosπ/6-cos2xsinπ/6-1/2=sin(2x-π/6)-1/2
-1<=sin(2x-π/6)<=1 -3/2<=sin(2x-π/6)-1/2<=1/2 值域为[-3/2,1/2]