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已知函数Fx=sinx四次幂+2倍根号三sinxCosx%Cosx的...

y=sin^4(x)+2√3(sinxcosx)-cos^4(x) =sin^4(x)-cos^4(x)+2√3sinxcosx =(sin^2x+cos^2x)(sin^2x-cos^2x)+2√3(sinxcosx) =2√3(sinxcosx)-cos2x =√3sin2x-cos2x =2sin(2x-π/6) 最小正周期T=2π/2=π 最小值是:2*(-1)=-2 单调增区间是:-Pai/2+2kPai

fx=2根号3sinxcosx-cos2x =2(根号3/2sin2x-1/2cos2x) =2sin(2x-派/6) 增区间: 2k派-派/3

fx=2sin²x+2√3sinxcosx+1 =1-cos2x+√3sin2x+1 =2sin(2x-π/6)+2 ∴周期是π,值域是:f(x)∈[0,4]

f(x)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

解1f(x)=cos2x+√3×2sinxcosx =cos2x+√3sin2x =2(1/2cos2x+√3/2sin2x) =2sin(2x+π/6) 故T=2π/2=π 当2kπ-π/2≤2x+π/6≤2kπ+π/2,k属于Z时,y是增函数。 即2kπ-2π/3≤2x≤2kπ+π/3,k属于Z时,y是增函数。 即kπ-π/3≤x≤kπ+π/6,k属于Z时,y是增函数。 故函数...

f(x)=2√3sinxcosx+2sin^2x-1=√3sin2x-cos2x=2sin(2x-π/6) 最小正周期T=π ,单调递增区间:2kπ-π/2

你好这个题目很简单就是利用正弦、余弦公式的变化加上函数周期性。原解:fx=1+2sinxcosx+(1+cos2x)-2 =sin2x+cos2x =√2 sin(2x+π/4) T=2π/2=π 望采纳

f(x)=2√3sinxcosx-2sin²x=√3sin2x+cos2x-1=2sin(2x+π/6)-1 1)T=2π/2=π 2)-π/6≤x≤π/4 -π/6≤2x+π/6≤2π/3 2x+π/6=π/2 x=π/6 f(-π/6)=-2 f(π/4)=√3-1 f(π/6)=1 ∴最大值为1,最小值为-2

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