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已知函数F(x)=4sinxsin²(π/4+x/2)+Cos2x%1

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

f'x = f'(pi/2) cosx -2sin2x ---1 f'(pi/2) = f'(pi/2) cos(pi/2) -2sinpi = f'(pi/2) *0 +2 所以f'(pi/2)=2 带入1)式子就得到f'x =2cosx-2sin2x 然后再把x=pi/4带入就得到了

(本小题满分14分)解:(1)f(x)=4sinx?1?cos(π2+x)2+cos2x?1=2sinx(1+sinx)-2sin2x=2sinx.∵f(ωx)=2sinωx在[?π2,2π3]是增函数,∴[?π2,2π3]?[?π2ω,π2ω]?2π3≤π2ω,∴ω∈(0,34](2)[12f(x)]2?mf(x)+m2+m?1=sin2x-2msinx+m2+m-1>0因为x∈[π...

已知函数f(x)=4SinX * Sin^2(π/4+x/2)+Cos2x (1)设W>0为常数,若Y=f(Wx)在区间[-π/2,2π/3]上时增函数,求W的范围!2)设集合A={X| π/6≤X≤2π/3},B={X|f(x)-m<2},若A∪B=B,求实数m的取值范围! 1) f(x)=4sinx*sin^2(π/4+x/2)+cos2x =...

解: (1) tanx有意义,x≠kπ+ π/2,(k∈Z) 函数定义域为{x|x≠kπ+ π/2,k∈Z} f(x)=4tanxsin(π/2 -x)cos(x- π/3) -√3 =4tanxcosxcos(x-π/3)-√3 =4sinx[cosxcos(π/3)+sinxsin(π/3)] -√3 =4sinx[(1/2)cosx+(√3/2)sinx] -√3 =2sinxcosx+2√3sin²x-√...

f(x)=4cosx(sinxcosπ/6+cosxsinπ/6)+1 =2√3sinxcosx+2cos²x-1+2 =√3sin2x+cos2x+2 =2(sin2x*√3/2+cos2x*1/2)+2 =2(sin2xcosπ/6+cos2xsinπ/6)+2 =2sin(2x+π/6)+2 所以T=2π/2=π -π/6

解:f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1 =-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1 =-√2(√2/2·sin2x+√2/2·cos2x)+3sin2x-cos2x =-sin2x-cos2x+3sin2x-cos2x =2sin2x-2cos2x =√[2²+(-2)²]sin(2x-π/4) =2√2sin(2...

f(X)=cos2X+sin2X =√2(sinπ/4cos2X+cosπ/4sin2X) =√2sin(2X+π/4) ∵X属于[0,π/2], ∴-√2/2≤sin(2X+π/4)≤1, ∴-1≤f(X)≤√2。 即当X=π/2时,f(X)最小=-1, 当X=π/4时,f(X)最大=√2。

(1)y=cos2x+sinxcosx=1+cos2x2+12sin2x=22 sin(2x+π4)+12∴T=2π2=π,由 2kπ?π2≤2x+π4≤π2+2kπ k∈Z,即 kπ?3π8≤x≤π8+kπ k∈Z,所以函数的单调增区间为:[?38π+kπ,π8+kπ] (k∈Z).(2)g(x)=2sin(x+π4)sin(x?π4)=-sin(2x+π2)=-cos2x,因为f(x...

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