y=2sin(sinx cosx)的最大值

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求函数y=sin²x+2sinxcosx+3cos²x的值域

y=sin²x+cos8x+2sinxcosx+2cos²x=1+sin2x+(1+cos2x)=sin2x+cos2x+2=√2sin(2x+π/4)+2-1

已知函数y=1-2sin^2x+2sinxcosx,求:

y=cos2x+sin2x=√2(√2/2*sin2x+√2/2cos2x)=√2(sin2xcosπ/4+cos2xsinπ/4)=√2sin(2x+π/4)所以T=2π/2=π值域是[-√2,√2

求函数y=cos²x-sin²x+2sinxcosx值域

y=cos²x-sin²x+2sinxcosx=cos2x+sin2x=√2(√2/2cos2x+√2/2sin2x)=√2sin(2x+π/4)所以可得此函数的值域为:[-√2,

y=sin²x+2sinxcosx+3cos²x ,x属于R

y=1+sin(2x)+2cos^2(x)=1+sin(2x)+1+cos(2x)=2+sin(2x)+cos(2x)=2+√2sin(2x+π/4)所以周期为π2.-π/2+2kπ

已知函数y=sin^2x+sinxcosx+2,(x属于R),求值域

y=(1-cos2x)/2+(sin2x)/2+2=1/2(sin2x-cos2x)+5/2=√2/2sin(2x-π/4)+5/2因此y的最大值为:√2/2+5/2,最小值为-√2/2+5/2值域即

已知函数y=2cosxsin(x+π/3)-根号3 *(sin^2) x +sinxcosx

y=2cosxsin(x+π/3)-根号3*(sin^2)x+sinxcosx,后两项先提出一个sinx,然后括号内部分用叠加原理,得到y=2cosxsin(x+π/3)+2sinxcos(x+π/3

求y=sin平方x+2sinxcosx的周期

y=sin^x+2sinxcosx=1/2-cos2x/2+sin2x=根号下(5/4)*[2sin2x/根号5-cos2x/根号5]+1/2设cosa=2/根号5,sina=-1/根号5上式=根号下

函数y=1/2sin(2X+TT/3)-sinXcosX的单调递减区间是

y=1/2sin(2X+TT/3)-sinXcosX=cos(2x+TT/6)sin(TT/6)=1/2cos(2x+TT/6)单调递减区间是2kπ≤2x+π/6≤2kπ+πkπ-π/12≤x≤kπ+

函数y=1/2sin(2x + pai/3)-sinxcosx的单调递减区间?

y=1/2sin(2x+π/3)-sinxcosx=1/2(1/2sin2x+√3/2cos2x)-1/2sin2x=1/2(-1/2sin2x+√3/2cos2x)=1/2(cosπ/6cos2x-

已知函数y=sin²x+2sinxcosx-3cos²x,x∈R

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

y=sin^2x-根号3sinxcosx-cos^2x化简

y=-(cos²x-sin²x)-√3sinxcosx=-cos2x-(√3/2)sin2x=-sin(2x+π/6)或者y=-(cos²x-sin²x)-√3

求函数y=sin^2x+根号3sinxcosx-1的值域

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

y=1/2cos^2x+sinxcosx+3/2sin^2x 求单调区间

hiy=1/2cos^2x+sinxcosx+3/2sin^2x=1/2*((1+cos2x)/2)+1/2*sin2x+3/2*((1+cos2x)/2)=1/2*sin2x-1/2*cos2x+1

求y=cos²x+2根号3sinxcosx-sin²x的最大值、最小值.

y=cos²x+2√3sinxcosx-sin²x=cos²x-sin²x+2√3sinxcosx=cos2x+2√3sinxcosx=cos2x+√3sin2

函数y=cos^X-sin^x+2sinxcosx的最小值是?

y=cos2x+sin2x=根号2*sin(2x+pai/4)故y(min)=-根号2

已知函数y=sin²x+sinxcosx+2(x∈R),求函数的值域

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

已知函数y=cos²x-sin²x+2sinxcosx,求函数值域

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

求函数y=7-8sinxcosx+4cos^2x-4sin^x的最小值

y=7-4sin2x+4cos2x=7+4√2cos(2x+π/4),y的最小值=7-4√2.