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函数F(x)=sin^2x+根号3sinxCosx在区间[兀/4,兀/2]...

max=3/2,min=1 解: f(x) =sin²x+√3sinxcosx =(1-cos2x)/2+(√3/2)sin2x =sin(2x-π/6)+1/2 ∵x∈[π/4,π/2] ∴2x-π/6∈[π/3,5π/6] ∴ f(x)_max=1+1/2=3/2 f(x)_min=1/2+1/2=1

第一个是sin2x还是sin²x?

sin2x=2sinxcosx 所以] √3sinxcosx= (√3 /2)*sin2x 不懂可以追问

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)=(1-cos2x)/2+√3/2*sin2x =√3/2sin2x-1/2*cos2x+1/2 =sin2xcosπ/6-cos2xsinπ/6+1/2 =sin(2x-π/6)+1/2 π/4

T=π; f(x)max=f(π/3)=3/2 f(x)=sin²x+√3sinxcosx =(1-cos2x)/2 + √3/2 sin2x =√3/2 sin2x - 1/2 cos2x + 1/2 =sin(2x-π/6) + 1/2 所以T=2π/2=π 因为,π/4≤x≤π/2,所以π/3≤2x-π/6≤5π/6 根据函数图像及性质可得,1/2≤sin(2x-π/6)≤1 所以1≤f(...

f( x)= sin2x +2根号3cosxsinx+sin(x-4分之派)sin(x+4分之派) f( x)= sin2x +根号3sin2x+sin(x-4分之派)cos(x-4分之派) f( x)= sin2x +根号3sin2x+1/2 sin(2x-2分之派) f( x)= (根号3+1)sin2x-1/2 cos2x

f(x)=sinxcosx+√3(cosx)^2-√3/2 =(1/2)sin2x+(√3/2)cos2x =sin2xcosπ/3+cos2xsinπ/3 =sin(2x+π/3) 1. 0

根号在哪啊

f(x)=(1-cos2x)/2+√3/2sin2x =1/2-cos2x/2+√3/2sin2x =1/2+sin(2x-30°) ∵45

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