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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(...

解fx=sinx的平方+根号3sinxcosx=(1-cos2x)/2+√3/2sin2x=√3/2sin2x-cos2x/2+1/2=sin(2x-π/6)+1/2故函数的周期T=2π/2=π.

f(x)=√3sinxcosx-cos²x+1/2 =√3/2*sin2x-1/2(2cos²x-1) =√3/2*sin2x-1/2*cos2x =sin(2x-π/6) (1)、最小正周期:T=2π/2=π 设sin(2x-π/6)=±1 则2x-π/6=π/2+kπ,k∈Z ∴对称轴方程为:x=π/3+kπ/2,k∈Z (2)、∵在△ABC中 f(A/2)=sin(A-π/6)=1/...

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]

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

f(x)=√3sinxcosx-1/2cos2x =√3/2sin2x-1/2cos2x =sin(2x-π/6). (1)最小值:f(x)|min=-1, 此时2x-π/6=2kπ-π/2, 即x=kπ-π/6 (k为整数); 最小正周期:T=2π/2=π. (2)f(C)=1,则 sin(2C-π/6)=1,即C=π/3. R=c/(2sinC)=√3/(2·√3/2)=1 (正...

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),...

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