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已知函数F(x)=2sin2x+23sinxCosx?1.(Ⅰ)求F(x...

(Ⅰ)f(x)=3sin2x?3sin2x?cos2x+2(sin2x+cos2x)=3sin2x+cos2x?sin2x=3sin2x+cos2x=2sin(2x+π6)∴f(x)的最大值是2.(Ⅱ)由条件得 sin(2A+C)=2sinA+2sinAcos(A+C),∴sinAcos(A+C)+cosAsin(A+C)=2sinA+2sinAcos(A+C),化简得 sinC=...

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1f(x)=cos2x+√3×2sinxcosx =cos2x+√3sin2x =2(1/2cos2x+√3/2sin2x) =2sin(2x+π/6) 故T=2π/2=π 当2kπ-π/2≤2x+π/6≤2kπ+π/2,k属于Z时,y是增函数。 即2kπ-2π/3≤2x≤2kπ+π/3,k属于Z时,y是增函数。 即kπ-π/3≤x≤kπ+π/6,k属于Z时,y是增函数。 故函数...

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

F(x)=|cos2x+sin2x+Ax+B|=|2sin(2x+π4)+Ax+B|,(1)若F(x)是周期函数,F(x+π)=F(x),即|2sin(2x+π4)+Ax+B|=|2sin(2π+2x+π4)+Ax+Aπ+B|,可得A=0,B为任意实数;(2)∵0≤x≤3π2,∴π4≤2x+π4≤13π4,∴-1≤2sin(2x+π4)≤2,当A=0,B=-2...

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+...

函数f(x)=2sinxcosx+23sin2x?3=sin2x+-3cos2x=2sin(2x-π3),y=f(x)的图象向左平移π6个单位,再向上平移1个单位,得到函数y=g(x)=2sin[2(x+π6)-π3]+1=2sin2x+1的图象,由题意可得,g(x)在[a,b]上至少含有1012个零点.令g(x)=0,得...

二倍角公式

f(x)=sin2x+√3sinxcosx+1/2 =sin2x+(√3/2)sin2x+1/2 =[(2+√3)/2]sin2x+1/2 (1) f(x)的最小正周期T=π (2) f(x)的最大值为(3+√3)/2 取得最大值时2x=2kπ+π/2(k∈Z) 所以取最大值时x的集合为{x|x=kπ+π/4,k∈Z} (3) f(x)的递减时,2x∈(2kπ+π/2,2kπ+3...

f(x)=sin²x+sinxcosx+cos2x =2分之(1-cos2x)+(2分之1)sin2x+cos2x =2分之1+(2分之1)(sin2x+cos2x) =(2分之√2)sin(2x+45°)+(2分之1) 最小正周期 = 2π÷2 = π

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