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设函数F(x)=23sinxCosx?2sin2x+1(x∈R),则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(...

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...

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

(1)f(x)=32sin2x-12cos2x=sin(2x-π6),∵ω=2,∴f(x)的最小正周期T=2π2=π,令2x-π6=kπ+π2,得到x=kπ2+π3(k∈Z),则图象的对称轴方程为x=kπ2+π3(k∈Z);(2)要得到函数g(x)=sinx的图象,只需将函数f(x)的图象首先向左平移π12,得到y...

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...

解答:(本小题满分13分)解:(1)f(x)=sinxcosx-12sin(2x-π3)=12sin2x-12(sin2xcosπ3-cos2xsinπ3)=12sin2x-14sin2x+34cos2x=12sin(2x+π3)则f(x)的最小正周期为π.…(7分)(2)因为x∈[0,π2],则2x+π3∈[π3,4π3].所以12sin(2x+π3...

已知函数f(x)=-(√2)sin(2x+π/4)+6sinxcosx-2cos²x+1,x∈R; (1).求f(x)的最小正周期;(2).求f(x)在区间[0,π/2]上的最大值和最小值. 解:(1)。f(x)=-(sin2x+cos2x)+3sin2x+cos2x=2sin2x,故最小正周期T=π; (2)。在区间[0,π/2]上的最大...

(1)y=2-sin2x+cos2x=?2sin(2x?π4)+2,∵2kπ?π2≤2x?π4≤2kπ+π2,∴kπ?π8≤x≤kπ+38π,∴该函数在[0,π]上的单调递减区间为[0,38π],[7π8,π].(4分)(2)T=π,由(1)问知:当x=78π+kπ,(k∈Z),倍f(x)最大值为2+2,当x=38π+kπ,(k∈Z),f(x)...

(1)∵sin2x=2sinxcosx∴当cosx>0时,即?π2+2kπ<x<π2+2kπ时,y=sin2x|cosx|=2sinxcosxcos=2sinx当cosx<0时,即π2+2kπ<x<3π2+2kπ时,y=sin2x|cosx|=2sinxcosx?cosx=?2sinx(2)∵y=sinx1+cosx1?cosx+|cosx|=sinx?1+cosx|sinx|+|cosx|∴当x...

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