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已知函数Fx=Cos2x%sin2x+4sinx·Cosx,求F(x)的最小...

(1)f(x)=√3sin2x+cos2x=2(√3/2sin2x+1/2cos2x)=2sin(2x+π/6) 所以T=2π/2=π 因为x∈[0,π/2] ==>2x+π/6∈[π/6,7π/6] 根据正弦曲线的单调性 f(x)max=2×1 当2x+π/6=π/2时取得 f(x)max=2×(-1/2)=-1 当2x+π/6=7π/6时取得 (2)由题意可得2sin(2x1+π/6)=...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

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(...

(1)f(x)=cos2x?sin2x+23sinxcosx+1=3sin2x+cos2x+1=2sin(2x+π6)+1.(4分)因此f(x)的最小正周期为π,最小值为-1.(6分)(2)由F(a)=2得2sin(2α+π6)+1=2,即2sin(2x+π6)=12,而由a∈[π4,π2],得2a+π6∈[23π,76π].(9分)故2a+π6=56...

f(x)=(cosx)^4-2sinxcosx-(sinx)^4 =[(cosx)^2+(sinx)^2][(cosx)^2-(sinx)^2]-2sinxcosx =(cosx)^2-(sinx)^2-2sinxcosx =cos(2x)-sin(2x) = √2*[cos(2x)*√2/2-sin(2x)*√2/2] = √2*[cos(2x)cos(π/4)-sin(2x)sin(π/4)] = √2cos(2x+π/4) 。 (2)因...

(Ⅰ)f(x)=cos2x+3sin2x=2sin(2x+π6),(6分)∴f(π12)=2sin(π6+π6)=2sinπ3=3.(8分)(Ⅱ)由(Ⅰ)可知f(x)=2sin(2x+π6),∴函数f(x)的最小正周期T=2π2=π.(11分)函数f(x)的最大值为2.(13分)

f(x)=cos2x+23sinxcosx-sin2x=(cos2x-sin2x)+3(2sinxcosx)=cos2x+3sin2x=2sin(2x+π6),(1)∵ω=2,∴T=2π2=π,又正弦函数的递增区间为[2kπ-π2,2kπ+π2](k∈Z),∴2kπ-π2≤2x+π6≤2kπ+π2(k∈Z),解得:kπ-π3≤x≤kπ+π6(k∈Z),则函数f(x)...

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。

f(x)=2cos2x+sin^x-4cosx =2(2cos^x-1)+(1-cos^x)-4cosx =3cos^x-4cosx-1 f(π/3)=3cos^(π/3)-4cosπ/3-1=3x(1/2)^-4x1/2-1=(-9)/4; 令t=cosx,t∈[-1,1]. F(t)=3(t^2)-4t-1 F'(t)=6t-4 令F'(t)=0,t=2/3; t∈[-1,2/3], F(t)递减,t∈[2/3,1],F(t)递增...

f(x)=2sinxcosx+cos2x =sin2x+cos2x =√2sin(2x+π/4) T=2π/2=π sin函数的单增区间为 -π/2+2kπ,π/2+2kπ k为整数 即 -π/2+2kπ ≤2x+π/4 ≤π/2+2kπ 即 -3π/8+kπ≤x≤π/8+kπ ∵x∈[0,π/3] ∴f(x)max时 即sin=1 即f(x)max=√2 f(x)min=f(π/3)=√2sin(11π/12)

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