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设函数F(x)=Cos^2x%根号3sinxCosx+1/2,(1)求F(x)...

解:(1)f(x)=cosx-√3sinxcosx+=[1+cos(2x)]-(√3/2)sin(2x)+=cos(2x)-(√3/2)sin(2x)+1=cos(2x+π/3)+1最小正周期T=2π/2=πcos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2cos(2x+π/3)=-1时,f(x)取得最小值f(x)min=-1+1=0函数的值

解1:f(x)=cosx-(√3)sinxcosx+1f(x)=(1/2)(2cosx-1)-[(√3)/2]2sinxcosx+3/2f(x)=(1/2)cos(2x)-[(√3)/2]sin(2x)+3/2f(x)=cos(π/3)cos(2x)-sin(π/3)sin(2x)+3/2f(x)=cos(2x+π/3)+3/2依余弦函数的性质,可知:当2kπ+π2kπ+π2kπ+2π/3解得:kπ+π/3即:f(x)的

f(x)=(1+cos2x)/2-√3sin2x/2+1 =cos(2x+π/3)+3/2(1)2kπ-π≤2x+π/3≤2kπ,k∈z所以闹递增区间为[kπ-2π/3,kπ-π/6],k∈z(2)f(a)=cos(2a+π/3)+3/2=5/6,所以cos(2a+π/3)=-2/3因为2a∈(2π/3,4π/3),所以2a+π/3∈(π,5π/3)sin(2a+π/3)=-√5/3sin2a=sin[(2a+π/3)-π/3]=(2√3-√5)/6

用二倍角公式啦 (cosx)^2 = (1+cos(2x))/2 根号(3) sinxcosx = 根号(3)/2 * sin(2x) 所以 f(x) = 1/2 * cos(2x) + 根号(3)/2 * sin(2x) +3/2 Sin(π/6) = 1/2 Cos(π/6) = 根号(3)/2 所以 f(x) = sin(π/6)cos(2x) + cos(π/6)sin(2x) + 3/2 = sin(2x + π/6) + 3/2 逆

1. 答:f(x)=√3sinxcosx-(cosx)^2+1/2=(√3sin2x)/2-(cos2x+1)/2+1/2=(√3sin2x-cos2x)/2=2sin(2x-π/6)/2=sin(2x-π/6) 所以f(x)的最小正周期T=2π/2=π2. 答:因为f(A/2+兀/3)=sin(A+2π/3-π/6)=sin(A+π/2)=cosA=1/3 所以sinA=2√2/3因为b=1 a=√2 所以利用正弦定理 所以a/sinA=b/sinB所以sinB=bsinA/a=(2√2/3)/√2=2/3

f(x)=根号3 cos^2x+sinxcosx - 根号3/2 =(√3/2)(1+cos2x)+(1/2)sin2x-√3/2=(√3/2)cos2x+(1/2)sin2x=sin(2x+π/3)(1) 最小正周期T=2π/2=π单增区间2x+π/3∈[2kπ-π/2, 2kπ+π/2]即x∈[kπ-5π/12, kπ+π/12](2) x∈[0, 3π] 2x+π/3∈[π/3, 6π+π/3]满足最大值条件

答:f(x)=(1/2)*(cosx)^2+(√3/2)sinxcosx+1=(1/2)*(cos2x+1)/2+(√3/4)sin2x+1=(1/2)[sin2xcosπ/6+cos2xsinπ/6]+5/4=sin(2x+π/6)/2+5/4因为:π/12

f(x)=(1+cos2x)/2-√3sin2x/2+1=cos(2x+π/3)+3/2(1)2kπ-π≤2x+π/3≤2kπ,k∈Z所以闹递增区间为[kπ-2π/3,kπ-π/6],k∈Z(2)f(a)=cos(2a+π/3)+3/2=5/6,所以cos(2a+π/3)=-2/3因为2a∈(2π/3,4π/3),所以2a+

解:f(x)=2cosx+2√3sinxcosx =(2cosx-1)+1+√3*2sinxcosx =cos2x+√3sin2x+1 =2sin(2x+π/6)+1 函数f(x)最小正周期T=2π/2=π 由2kπ≤2x+π/6≤2kπ+π/2 (k∈z),得 kπ-π/12≤x≤kπ+π/6 或 由2kπ+3π/2≤2x+π/6≤2kπ+2π(k∈z),得 kπ+2π/3≤x≤kπ+11π/12 因此,f(x)的单独递增区间为[kπ-π/12,kπ+π/6]或[kπ+2π/3,kπ+11π/12] (k∈z)

f(x)=2cos^2x +2√3sinxcosx +1 =1+cos2x+√3sin2x+1 =2+2(√3/2*sin2x+1/2*cos2x) =2+2sin(2x+π/6)由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z得 kπ-π/3≤x≤kπ+π/6∴f(x)单调增区间[ kπ-π/3,kπ+π/6],(k∈Z)

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