已知函数y=sin²x-1 2sinx 1,若当y取最大值-搜答案-拍题作业网

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

设t=3x+π/3,则y=sin(3x+π/3)=sint的单调递增区间为：2kπ-π/2≤t≤2kπ+π/2,k∈Z也即2kπ-π/2≤3x+π/3≤2kπ+π/2得2kπ/3-5π/18≤x≤2k

两边求导得：cos(xy)*(y+xy')+1+y'=0y'[xcos(xy)+1]=-ycos(xy)-1所以,y'=-[ycos(xy)+1]/[xcos(xy)+1]

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=

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

方程y=sin(x+y)两边对x求导数有：y'=cos(x+y)(x+y)'=cos(x+y)(1+y')移项整理得：[1-cos(x+y)]y'=cos(x+y)因此：y'=cos(x+y)/[1-

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

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