Question

求函数y=sin^2x+根号3sinxcosx-1的值域

Answer 1

y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2 +(√3/2)sin(2x) -1=(√3/2)sin(2x) -(1/2)cos(2x) -1/2=sin(2x -π/6) -1/2-1≤sin(2x-π/6)≤1-3/2≤sin(2x-π/6) -1/2≤1/2-3/2≤y≤1/2函数的值域为[-3/2,1/2].

Answer 2

y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2 +(√3/2)sin(2x) -1=(√3/2)sin(2x) -(1/2)cos(2x) -1/2=sin(2x -π/6) -1/2-1≤sin(2x-π/6)≤1-3/2≤sin(2x-π/6) -1/2≤1/2-3/2≤y≤1/2函数的值域为[-3/2,1/2].

Answer 3

y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2 +(√3/2)sin(2x) -1=(√3/2)sin(2x) -(1/2)cos(2x) -1/2=sin(2x -π/6) -1/2-1≤sin(2x-π/6)≤1-3/2≤sin(2x-π/6) -1/2≤1/2-3/2≤y≤1/2函数的值域为[-3/2,1/2].

Answer 4

y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2 +(√3/2)sin(2x) -1=(√3/2)sin(2x) -(1/2)cos(2x) -1/2=sin(2x -π/6) -1/2-1≤sin(2x-π/6)≤1-3/2≤sin(2x-π/6) -1/2≤1/2-3/2≤y≤1/2函数的值域为[-3/2,1/2].

Answer 5

y=sin²x+√3sinxcosx-1=[1-cos(2x)]/2 +(√3/2)sin(2x) -1=(√3/2)sin(2x) -(1/2)cos(2x) -1/2=sin(2x -π/6) -1/2-1≤sin(2x-π/6)≤1-3/2≤sin(2x-π/6) -1/2≤1/2-3/2≤y≤1/2函数的值域为[-3/2,1/2].