已知函数y=1 2sin^2x 根3 2sinxcosx-搜答案-智慧作业帮

振幅为2；周期为π；初相为π/3单增区间：kπ－5π／12≦x≦kπ＋π／12对称轴：x=﹙1／2﹚kπ+（1／12）π

（0,1）代入原式知sinφ=1/2因为|φ|

解由x属于[π/12,π/3]即π/12≤x≤π/3即π/6≤2x≤2π/3即-π/6≤2x-π/3≤π/3即-1/2≤sin（2x-π/3）≤√3/2即-1≤2sin（2x-π/3）≤√3即-1≤y

f(x)=0.5(1-cos2x)+0.5sin2x+cos2x=0.5+0.5(sin2x+cos2x)=0.5+0.5根号2sin(2x+π/4)f(x)的最小值是（1-根号2)/2

由化简sinx+cosx前分别乘以根号2*sin45.根号2*cos45.,得解根号2sinxy=sinx的平方+根好2*sinx+2令t=sinx-1=

如果是y=sin2x+sinx+cosx+2,那好求,y=1+2sinxcosx+(sinx+cosx)+1=(sinx+cosx)²+(sinx+cosx)+1令t=sinx+cosx=√

y=sin²x+sinx+cosx+2=(1-cos2x)/2+√2sin(x+л/4)+2=(1/2)*sin(2x+л/2)+√2*sin(x+л/4)+5/2；=(1/2)*sin(2

我列个去,就算我高中毕业到现在已经8年了,我也看的出来1楼的乱说的撒,值域明显是[-2,2]嘛

(-π/2,π/2)应小于等于半个周期,.-1≤ω≤1,又函数是减函数,sin(-ωπ/2)>sin(ωπ/2),sin(ωπ/2)

（1）原式=2[(1/2)sin(x/2)+（根号3/2）cos(x/2)]=2sin[(x/2)+pi/3]所以当[(x/2)+pi/3]=2kpi+pi/2时,y最大值为2解得x=4kpi+pi/

y=sin²x+2sinxcosx-3cos²x=(sin²x+cos²x)+2sinxcosx-4cos²x=1+sin(2x)-2[1+cos(2

因为,-π/2

解由y=sin（pai/4-2x）=-sin(2x-π/4)知当2kπ-π/2≤2x-π/4≤2kπ+π/2,k属于Z时,y是减函数.即当kπ-π/8≤x≤kπ+3π/8,k属于Z时,y是减函数.故函

y=cos2x+sin²x-cosx=cos²x-cosx=(cosx-1/2)²-1/4x=2kπ+π,max(y)=2x=2kπ±π/3,min(y)=-1/4x∈[

y=sin²x+sinxcosx+2=(1-cos2x)/2+(sin2x)/2+2=(1/2)(sin2x-cos2x)+5/2=(1/2)*√2(sin2xcosπ/4-cos2xsin

原式=(1-cos2x)/2+(sin2x)/2+2=(sin2x-cos2x)/2+5/2=(sin（2x-45度）)*(根号2)/2+5/2所以是大于（根号2+5）/2,小于（5-根号2）/2

解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数

函数y=2sin(3x+π/6)当函数y取最大值时有3x+π/6=2kπ+π/2即x=2kπ/3+π/9,k∈Z所以x得集合为{x|x=2kπ/3+π/9,k∈Z}

y=cos²x-sin²x+2sinxcosx=cos2x+sin2x=√2sin(2x+π/4)所以值域为【-√2,√2】