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已知函数F(x)=Cos^4x%2sinxCosx%sin^4x

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 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

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.

原式=(cos^2x+sin^2x)(cos^2x-sin^2x)-2sinxcosx=cos^2x-sin^2x-2sinxcosx=cos2x-sin2x=√2[(√2/2)cos2x-(√2/2)sin2x]=√2cos(π/4+2x)周期2π/2=π当π/4+2x=π+2kπ时,即x=3π/8+kπ时,f(x)最小=-√2

原式=(cos^2x+sin^2x)(cos^2x-sin^2x)-2sinxcosx=cos^2x-sin^2x-2sinxcosx=cos2x-sin2x=√2[(√2/2)cos2x-(√2/2)sin2x]=√2cos(π/4+2x)周期2π/2=π当π/4+2x=π+2kπ时,即x=3π/8+kπ时,f(x)最小=-√2

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

f(x)=cos^4x-2sinxcosx-sin^4x=(cosx+sinx)(cosx-sinx)-2sinxcosx=1*cos2x-sin2x;=√2(cos2x*(√2/2)-sin2x*(√2/2))=√2sin(2x-π/4)如果本题有什么不明白可以追问,如果满意记得采纳如果有其他问题请采纳本题后另发点击向我求助,答题不易,请谅解,谢谢.祝学习进步

f(x)=coa^4x-2sinxcosx-sin^4x=(coa^2x+sin^2x)(coa^2x-sin^2x)-2sinxcosx=coa^2x-sin^2x-2sinxcosx=cos2x-sin2x=√2(√2/2*cos2x-√2/2sin2x)=√2(cos2xcosπ/4-sin2xsinπ/4)=√2cos(2x+π/4)所以T=2π/2=π

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

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