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已知函数f(x)=cos^4x-2sinxcosx-sin^4x

f(x)=cos^4x-2sinxcosx-sin^4x 则f(x)=(cos^2x+sin^2x)(cos^2x-sin^2x)-2sinxcosx =cos^2x-sin^2x-2sinxcosx =cos2x-sin2x =根号2*cos(2x+兀/4)则T=2兀/2=兀 当x ∈[0,π/2],则2x+π/4∈[π/4,5π/4] 则f(x)∈[-根号2,1] 当f(x)=-根号2时 x=3π/8 当f(x)=1时 x=0

f(x)=(cosx+sinx)(cosx-sinx)-sin2x=cos2x-sin2x=-(√2)sin[2x-(π/4).即f(x)=-(√2)sin[2x-(π/4)].∴(1)T=π.(2)2kπ-(π/2)≤2x-(π/4)≤2kπ+(π/2).===>kπ-(π/8)≤x≤kπ+(3π/8).∴

f(x)=cos^4x-2sinxcosx-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)-2sinxcosx=cos2x-sin2x=√2cos(2x+π/4)所以值域为[-√2,√2]

f(x)=cos^4x-2sinxcosx-sin^4x则f(x)=(cos^2x+sin^2x)(cos^2x-sin^2x)-2sinxcosx=cos^2x-sin^2x-2sinxcosx=cos2x-sin2x=根号2*cos(2x+兀/4)则T=2兀/2=兀 当x ∈[0,π/2],则2x+π/4∈[π/4,5π/4] 则f(x)∈[-根号2,1] 当f(x)=-根号2时 x=3π/8 当f(x)=1时 x=0

f(x)=cos^4x-sin^4x-sin2x=cos^2x-sin^2x-sin2x=cos2x-sin2x=√2sin(2x-pi/4)所以f(x)的最大值是√2最小值为-√22)先把sinx向右平移pi/8个单位,周期缩小2倍,振幅扩大√2倍

1.原式=(cos^x+sin^x)(cos^x-sin^x)-2sinxcosx=(cos^x-sin^x)-sin2x=cos2x-sin2x=根号2sin(π/4-2x)T=2π/(π/4)=82.由题,2kπ-π/2

f(x)=(cosx)^4-(sinx)^4-2sinxcosx=[(cosx)^2-(sinx)^2]*[(cosx)^2+(sinx)^2]-2sinxcosx=(cosx)^2-(sinx)^2-2sinxcosx=cos2x-sin2x=√2cos(2x+π/4)x∈[0,π/2]则2x+π/4∈[π/4,5π/4]最大值1 此时2x+π/4=π/4 x=0

f(x)=cos^4x-2sinxcosx-sin^4x =(cosx)^2-(sinx)^2-2sinxcosx=cos2x-sin2x=(√2)cos(2x+π/4),(1)f(x)的最小正周期是π.(2)x∈[0,π/2],∴2x+π/4∈[π/4,5π/4],∴f(x)的最小值是-√2,这时2x+π/4=π,x=3π/8.

答:f(x)=(cosx)^4-2sinxcosx-(sinx)^4=(cosx-sinx)(cosx+sinx)-sin2x=cos2x-sin2x=√2 [(√2/2)cos2x-(√2/2)sin2x ]=√2cos(2x+π/4)(1)f(x)的最小正周期T=2π/2=π;最大值为√2(2)0<x<π/2,π/4<2x+π/4<5π/4所以:-1<=cos(2x+π/4)<√2/2所以:f(x)的最小值为-√2当2x+π/4=π即x=3π/8时取得最小值

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