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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

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

f(x)=sin²x+√3sinxcosx =(1/2)(1-cos2x)+(√3/2)sin2x =(√3/2)sin2x-(1/2)cos2x+1/2 =sin(2x-π/6)+1/2 因为 x∈[π/4,π/2] 所以 2x-π/6∈[π/3,5π/6] 所以 sin(2x-π/6)∈[1/2,1] 的f(x)的最大值为 f(π/3)=1+1/2=3/2

应该是 f(x)=sin^2x+√3sinxcosx=-(-2sin^2x\2)+(√3\2)sin2x =-[(1-2sin^2x)\2]+(√3\2)sin2x+1\2 =-(1\2)cos2x+(√3\2)sin2x+1\2 =sin(2x-π\3)+1\2 所以当2x-π\3=π\2 x=5π\12 时最大值为1+1\2=3\2

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

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

公式cos 2x=1-2sin²x,可以知道sin ²x=(1-cos 2x)/2 后面的√3sin x cosx =√3sin 2x/2 所以原式=- (cos 2x)/2+(√3sin2x)/2+1/2 =sin(2x-π/6)+1/2 x属于[π/4,π/2],所以2x+π/6属于[π/3,5π/6].就是π/2是可以取到的,前面三角函数最大值为...

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)(1/2)+(√3)/2sin2x-1【三角变形】 =(√3)/2sin2x-(1/2)cos2x-1 =sin(2x+π/6)-1 f(x)的最小正周期:T=2π/w (w=2)=π  当x= π/6时,f(x)最大=0,当x= π /2时,f(x)的最小=-1/2 f(x)的值域为【-1/2,0】

Y=sin²x+√3sinxcosx =(1-cos(2x))/2+√3/2*sin(2x) =√3/2*sin(2x) -cos(2x)/2+1/2 = sin(2x-π/6) +1/2 X∈[π/4, π/2], 2x-π/6∈[π/3, 5π/6], 当2x-π/6=5π/6时,函数取到最小值1/2+1/2=1. 当2x-π/6=π/2时,函数取到最大值1+1/2=3/2.

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