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寻求函数F(x)=sin平方x+(根号3)sinxCosx 在区间[π/4,π/2]上的最大值是多少...

已知函数F(X)=SIN平方X+根号3SINXCOSX+2COS平方X,求函数F(X)的最小正周期和单调递增区间 f(x)= (sinx)^2 +3 cosx*sinx + 2(cosx)^2 = (cosx)^2+1 + 1.5*2sinx*cosx = -cos2x/2 +1.5 +1.5 sin2x 展开 作业帮用户 2017-09-20 举报

f(x)=(1-cos2x)/2+根号3/2sin2x=1/2+sin(2x-π/6)x∈【π/4,π/2】,2x-π/6∈【π/3,5π/6】,sin(2x-π/6)∈【1/2,1]f(x)max=f(π/3)=3/2f(x)min=f(π/2)=1

最大值是:sinA+(√3)/2解:f(x)=sinA+(√3)sinxcosxf(x)=sinA+[(√3)/2]sin(2x)f'(x)=(√3)cos(2x)因为:x∈[π/4,π/2]所以:2x∈[π/2,π]显然,有:f'(x)所以,有:f(x)的最大值是:f(π/4)=sinA+(√3)sin(π/4)cos(π/4)=sinA+(√3)/2

应该是f(x)=sin^2x+√3sinxcosx=-(-2sin^2x\2)+(√3\2)sin2x =-[(1-2sin^2x)\2]+(√3\2)sin2x+1\2 =-(1\2)cos2x+(√3\2)sin2x+1\2 =sin(2x-π\3)+1\2所以当2x-π\3=π\2 x=5π\12 时最大值为1+1\2=3\2

f(x)=sin 2 x+ 3 sin xcos x= 1-cos2x 2 + 3 2 sin 2x=sin(2x- π 6 )+ 1 2 .∵ π 4 ≤x≤ π 2 ,∴ π 3 ≤2x- π 6 ≤ 5 6 π.当sin(2x- π 6 )=1,即2x- π 6 = π 2 时,此时x= π 3 ,函数f(x)取到最大值

答案:3/2f(x)=(sinx)^2+√3sinaxcosx =(sinx)^2+√3/2*2sinaxcosx(2sinaxcosx可用两倍角公式) =(sinx)^2+√3/2*sin2x =(1-cos2x)/2+√3/2*sin2x(因为cos2x=1-2(sinx)^2,所以(sinx)^2=(1-cos2x)/2,所以得出上式) =1/2-cos2x/2+√3/2*sin2x =1/2*(√3sin2x-cos2x+1) =1/2*[2sin(2x-pi/6)+1] (辅助角公式)当x=pi/3时,f(x)max=3/2

f(x)=sinx+√3sinxcosx=(1-cos2x)/2+(√3/2)sin2x=(√3/2)sin2x-(1/2)cos2x+1/2=sin(2x-π/6)+1/2x∈[π/4,π/2]π/3≤2x-π/6≤5π/6所以f(x)的最大值是1+1/2=3/2

f(x)=sin^2x+√3sinxcosx f(x)=[(1-cos2x)/2]+√3sin2x/2 f(x)=1/2+√3sin2x/2-cos2x/2 f(x)=1/2+√[(√3/2)^2+(1/2)^2]sin(2x-π/6) f(x)=1/2+sin(2x-π/6) π/4<=x<=π/2 π/3<=2x-π/6<=5π/6 所以f(x)max=3/2

解:y=sinx+√3sinxcosx =(1-cos2x)/2+√3/2sin2x =√3/2sin2x-1/2cos2x+1/2 =cosπ/6sin2x-sinπ/6cos2x+1/2 =sin(2x-π/6)+1/2因为π/4≤x≤π/2所以π/2≤2x≤ππ/2-π/6≤2x-π/6≤π-π/6 π/3≤2x-π/6≤5π/6所以当2x-π/6=π/2,时y取得最大值,最大值为y=1+1/2=3/2答案:3/2

将函数降次得:f(x)=(1-cos2x)/2+√3sin2x/2=1/2+3sin2x/2-cos2x/2=sin(2x-π/6)+1/2又:x∈[π/4,π/2],则(2x-π/6)∈[π/3,5π/6],在此区间,(2x-π/6)取5π/6时,f(x)最小,此时sin(2x-π/6)值为1/2,所以所求最小值为1.

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