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设函数F(x)=2根号3sinxCosx+2Cos^2x+m(1)求函数F...

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(...

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)=√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。

=√3sin2x-sin方x+1/2cos2x+1/2 =√3sin2x-(1-cos2x)/2+1/2cos2x+1/2 =2sin(2x+∏/6) -∏/4

解: f(x)=sin²x-cos²x-2√3sinxcosx =-(cos²x-sin²x)-√3(2sinxcosx) =-cos2x-√3sin2x =-2[(√3/2)sin2x+(1/2)cos2x] =-2sin(2x+π/6) (1) f(2x/3)=-2sin[2(2x/3)+ π/6]=-2sin(4x/3 +π/6) (2) 最小正周期T=2π/2=π 2kπ+ π/2≤2x+...

f(x)=2根号3sinxcosx+2cos^2x-1 =√3 sin2x+cos2x =2sin(2x+π/6) f(A)=2sin(2A+π/6)=2/3 sin(2A+π/6)=1/3 0

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]

根号的括号到哪里啊? 要是到(3sinXcosX+2cos^2X)和到 根号(3sinXcosX+2cos^2X-1)。要是后一种结果简单一点: 根号里面化简得到 [3/2sin(2X)+cos(2X)]=(根号13)/2 X sin(2X+φ),其中sinφ=2/(根号13),cosφ=3/(根号13) 则sin(2X+φ)的单调...

f(x)=2√3sinxcosx+2cos^2x =√3*sin2x+cos2x+1 =2(√3/2*sin2x+1/2*cos2x)+1 =2sin(2x+π/6)+1 f(x)的最小正周期T=2π/2=π f(x)min=-2+1=-1,f(x)max=2+1=3 f(x)值域为[-1,3]

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