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高二数学:已知函数F(x)=√3sinXCosX%Cos^2X%1/2...

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: (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(...

1:∵函数f(x) =√3sinxcosx-(cosx)^2-1/2 =cosx(√3sinx-cosx)-1/2=2cosxsin(x-π/6)-1/2 =sin(2x-π/6)-sin(π/6)-1/2 =sin(2x-π/6)-1 ∴f(x)的最小值为-2,最小正周期为π 2。f(C)=sin(2c-π/6)-1=0 则sin(2c-π/6)=1 当2c-π/6=π/2时。f(c)=0求...

f(x)=√3sinxcosx-cos²x-1/2 =(√3/2)sin2x-(1+cos2x)/2-1/2 =(√3/2)sin2x-(1/2)cos2x-1 =sin(2x-π/6)-1. 早小正周期T=2π/2=π. sin(2x-π/6)=-1, 即x=kπ-π/6时, 最小值f(x)|min=-2。

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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...

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

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 希望能帮助你:

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