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已知函数F(x)=2sin2x+23sinxCosx?1.(Ⅰ)求F(x...

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

依题意得:f(x)=cos2x-sin2x+23sinxcosx=cos2x+3sin2x=2sin(2x+π6),(1)∵x∈R,∴f(x)max=M=2,最小正周期T=2π2=π;(2)由f(xi)=M=2得:2xi+π6=2kπ+π2,k∈Z,解得:xi=kπ+π6,k∈Z,又0<xi<10π,∴k=0,1,2,…,9,∴x1+x2+…+x10=(1+2...

(sinx-cosx)sinx =(sinx)^2-sinxcosx =[-1+2(sinx)^2]/2+1/2-(sin2x)/2 =(-1/2)(cos2x+sin2x)+1/2 =(-√2/2)sin(2x+φ)+1/2 =(-√2/2)sin[2(x+φ/2)]+1/2 ∴最小正周期是(2π)/2=π

fx=4√3sinxcosx-4sin2x+1 =2√3sin2x-4sin2x+1 =(2√3-4)sin2x+1 f'x=(4√3-8)cos2x 单调增区间:f'(x)>0 ∴cos2x

解: f(x)=2√3sinxcosx+1 -2sin²x =√3sin(2x)+cos(2x) =2[(√3/2)sin(2x)+(1/2)cos(2x)] =2sin(2x+π/6) -1≤sin(2x+π/6)≤1 -2≤2sin(2x+π/6)≤2 -2≤f(x)≤2 函数的值域为[-2,2]

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

f(x)=2cosxcos(x?π6)?3sin2x+sinxcosx=2cosx(32cosx+12sinx)-3sin2x+sinxcosx=3(cos2x-sin2x)+2sinxcosx=3cos2x+sin2x=2(32cos2x+12sin2x)=2sin(2x+π3),(1)∵ω=2,∴T=2π2=π;(2)∵f(x)=1,即2sin(2x+π3)=1,∴sin(2x+π3)=12,...

f(x)=2√3sinxcosx-2sin²x+1 =√3sin2x+cos2x =2sin(2x+π/6) (2) 2kπ-π/2

看来你是看到我以前的回答了。那就直接复制一下吧。 解答: (1)sinx≠0, x≠kπ,k∈Z f(x)=(sinx-cosx)sin2x/sinx =(sinx-cosx)*2cosx =2sinxcosx-2cos²x =sin2x-cos2x-1 =√2sin(2x-π/4)-1 T=2π/2=π (2)增区间为 2kπ-π/2≤2x-π/4

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