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已知函数 F(x)=Cos2x+2 3 sinxCosx(x∈...

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...

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(...

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)=2cosxcos(x?π6)?3sin2x+sinxcosx=2cosx(32cosx+12sinx)-3sin2x+sinxcosx=3(cos2x-sin2x)+2sinxcosx=3cos2x+sin2x=2(32cos2x+12sin2x)=2sin(2x+π3),(1)∵ω=2,∴T=2π2=π;(2)∵f(x)=1,即2sin(2x+π3)=1,∴sin(2x+π3)=12,...

(Ⅰ)f(x)=2sinxcosx+2cos2x=sin2x+cos2x+1=2sin(2x+π4)+1,由2kπ-π2≤2x+π4≤2kπ+π2(k∈Z)得:kπ-3π8≤x≤kπ+π8,k∈Z,∴f(x)的单调递增区间是[kπ-3π8,kπ+π8](k∈Z);(Ⅱ)由题意得:g(x)=2sin(2x-π4)+1,由A(0,-1),得2sin(2α-π4...

(本小题满分15分)解:(Ⅰ)∵f(x)=32sin2x+1+cos2x2+a=sin(2x+π6)+12+a,∴最小正周期T=2π2=π,单调递减区间为[kπ+π6,kπ+2π3](k∈Z).(Ⅱ)令u=2x+π6,则g(u)=sinu+12+a,u∈[π6,7π6].要使g(u)在[π6,7π6]上恰有两个x的值满足g(u)=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+...

答: f(x)=(sinx)^2-2sinxcosx+3(cosx)^2 =(sinx)^2+(cosx)^2-sin2x+2(cosx)^2 =1-sin2x+cos2x+1 =cos2x-sin2x+2 =√2*[(√2/2)cos2x-(√2/2)sin2x]+2 =√2*cos(2x+π/4)+2 最小正周期T=2π/2=π

(Ⅰ)f(x)=3cos2x+2sin(3π2+x)sin(π-x)=3cos2x-2cosxsinx=3cos2x-sin2x=2(32cos2x-12sin2x)=2cos(2x+π6),∴T=2π2=π,令2x+π6=kπ(k∈Z),即x=kπ2-π12(k∈Z),∴函数f(x)的对称轴方程为x=kπ2-π12(k∈Z),(Ⅱ)∵f(x)=2cos(2x+π6)...

(Ⅰ)g(x)=cos2x+3sinxcosx?12=1+cos2x2+32sin2x?12…(2分)=12cos2x+32sin2x=sin(2x+π6)…(3分)∴函数g(x)的图象向右平移π12个单位,得g(x+π12)=sin2x,再将横坐标伸长到原来的2倍(纵坐标不变),得h(x)=sinx,…(4分)再将函数h(x...

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