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已知函数y=sin2x+sinx+Cosx+2,(x∈R),求函数y的...

y=sin2x+sinx+cosx+2 y=2sinxcosx+1+sinx+cosx+1 y=[2sinxcosx+sin^2(x)+cos^2(x)]+sinx+cosx+1 y=(sinx+cosx)^2+(sinx+cosx)+1 y=(sinx+cosx+1/2)^2+3/4 又因为:负根号2

y=sinx+sin2x-cosx =(sinx-cosx)+cos(2x- π/2) =√2[(√2/2)sinx-(√2/2)cosx]+cos[2(x- π/4)] =√2sin(x-π/4)+1-2sin²(x-π/4) =-2[sin(x-π/4) -√2/4]² +5/4 你的题目不全,(x∈),没有给出x的取值范围 不过,上述解题过程已经配方完成,...

(1)∵sin2x=2sinxcosx∴当cosx>0时,即?π2+2kπ<x<π2+2kπ时,y=sin2x|cosx|=2sinxcosxcos=2sinx当cosx<0时,即π2+2kπ<x<3π2+2kπ时,y=sin2x|cosx|=2sinxcosx?cosx=?2sinx(2)∵y=sinx1+cosx1?cosx+|cosx|=sinx?1+cosx|sinx|+|cosx|∴当x...

令f(x)=sin2x+2sinx f'(x)=2cos2x+2cosx=4(cosx)^2+2cosx-2=2(2cosx-1)(cosx+1) 当f'(x)=0时,f(x)存在极值 此时,2cosx-1=0 cosx=1/2 则:sinx= (根号3)/2,或 -(根号3)/2 则:sin2x=2sinx*cosx=(根号3)/2,或-(根号3)/2 f(x)极值 = (根号3)/2 ...

2sinxcosx=sin(2x) cos²x-sin²x=cos(2x) y=cos(2x)+sin(2x) =根号2【sin(2x+π/4)】 所以最小正周期为π,最大最小值分别为正负根号2

∵sin2x-sinx+12=(sinx?12)2+14,由sinx∈[-1,1]得sin2x-sinx+12∈[14,52],根据指数函数y=(12)x是定义域上的减函数,∴函数的值域是[2?52,2?14],故答案是[2?52,2?14].

f(x)=sin^2x+√3sinxcosx+2cos^2x =cos^2x+√3/2*2sinxcosx+1 =1/2cos2x+√3/2sin2x+3/2 =sinx(2x+π/6)+3/2 T=2π/2=π 2x+π/6在[2kπ-π/2,2kπ+π/2]上单调递增 x在[kπ-5π/12,kπ+π/6]上单调递增 2)y=sin2x的图像经X轴向左平移π/12个单位得到y=sinx(2x+...

令t=sinx,则-1≤t≤1y=sin2x+sinx=t2+t=(t+12)2-14;∴函数在[-1,-12]上单调减,在[-12,1]上单调增∴t=-12时,函数取得最小值为-14;t=1时,函数取确定最大值2.即函数y=sin2x+sinx的最大值为:2.故选:D.

(1)保证tanx及tan(x/2)有意义,有x≠kπ+π/2,x/2≠kπ+π/2,联立解得x≠kπ+π/2且x≠2kπ+π (k∈Z) (2)y=sin2xtanx+sinxtan(x/2) =2sinxcosxsinx/cosx+sinx[(1-cos)/sinx] =2sin2x+1-cos =2(1-cos2x)+1-cosx = -2cos2x-cosx+3 =(25/8)-2(cosx+1/4)2 当co...

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