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已知函数Fx)=根号3sinxCosx%Cosx^2%1\2求函数的最...

积化和差与和差化积 f(x)=3/2sin2x-1/2(cos2x-1)-1/2=√10/2( 3/√10 sin2x-1/√10 cos2x)=√10/2sin(2x-α)然后都是书本知识应用了。

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(...

fx=2根号3sinxcosx-cos2x =2(根号3/2sin2x-1/2cos2x) =2sin(2x-派/6) 增区间: 2k派-派/3

解 f(x)=√3cos²x+sinxcosx-√3/2 =√3*(1+cos2x)/2+(1/2)sin2x-√3/2 =(1/2)sin2x+(√3/2)cos2x =sin(2x+π/3) ∴T=π 单增区间: -π/2+2kπ≤2x+π/3≤π/2+2kπ,k∈Z -5π/6+2kπ≤2x≤π/6+2kπ,k∈Z -5π/12+kπ≤x≤π/12+2kπ,k∈Z 即为:[-π/12+kπ,5π/12+kπ],k∈Z

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

求采纳,

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

你好:f(x)=√3sinxcosx–cos²x+1/2 =√3sinxcosx–1/2(1-cos²x)+1/2 =√3/2sin2x+1/2cos²x-1/2+1/2 =sin(2x+π/6)T=2π/2=π f(x)=√3sinxcosx-cos²x+1/2=(√3/2)(2sinxcosx)-(1/2)(2cos²x-1) 二倍角公式:2sinxcosx=sin(2x),...

fx=cosx的平方+根号3sinxcosx+2sinxcos(x+30°) =cosx的平方+根号3sinxcosx+2sinx(根号3/2cosx-1/2sinx) =cosx的平方+根号3sinxcosx+根号3sinxcosx-sinx平方 =2cosx的平方-1+2根号3sinxcosx =cos2x+根号3sin2x =2*(1/2cos2x+根号3/2sin2x) =2si...

(1).f(x) =√3/2sin2x+1/2cos2x =sin(2x+π/6) T=2π/ω=π (2)将y=sinx的图像向右平移π/6个单位得y=sin(x+π/6),再将y=sin(x+π/6)的图像上的点纵坐标不变,横坐标缩短一倍得y=sin(2x+π/6).

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