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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

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)=√3sinxcosx+cos²x =(√3/2)·2sinxcosx+½(2cos²x-1)+½ =(√3/2)sin2x+½cos2x+½ =sin(2x+π/6) +½ 最小正周期T=2π/2=π

=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.

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)=sin^2x+√3sinxcosx+2cos^2x =cos^2x+√3/2*2sinxcosx+1 =1/2cos2x+√3/2sin2x+3/2 =sinx(2x+π/6)+3/2 T=2π/2=π 2x+π/6在[2kπ-π/2,2kπ+π/2]上单调递增 x在[kπ-5π/12,kπ+π/6]上单调递增 2)y=sin2x的图像经X轴向左平移π/12个单位得到y=sinx(2x+...

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

2x+π/3∈(π/3,4π/3) 2x+π/3看成一个整体a.取值a∈(π/3,4π/3) sin a (-√3/2,1]就是个sin的函数. 所以sin(2x+π/3)∈(-√3/2,1]

fx=4√3sinxcosx-4sin2x+1 =2√3sin2x-4sin2x+1 =(2√3-4)sin2x+1 f'x=(4√3-8)cos2x 单调增区间:f'(x)>0 ∴cos2x

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...

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