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

(1)f(x)=2sin2x+23sinxcosx?1=3sin2x?cos2x=2sin(2x?π6),则T=2π2=π,(2)当(2x?π6)=3π2+2kπ(k∈Z)时,sin(2x?π6)=?1,则函数f(x)取得最小值为-2.此时,x=5π6+kπ(k∈Z).

(1)f(x)=cos2x?sin2x+23sinxcosx+1=3sin2x+cos2x+1=2sin(2x+π6)+1.(4分)因此f(x)的最小正周期为π,最小值为-1.(6分)(2)由F(a)=2得2sin(2α+π6)+1=2,即2sin(2x+π6)=12,而由a∈[π4,π2],得2a+π6∈[23π,76π].(9分)故2a+π6=56...

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。

(Ⅰ)f(x)=cos2x+3sin2x=2sin(2x+π6),(6分)∴f(π12)=2sin(π6+π6)=2sinπ3=3.(8分)(Ⅱ)由(Ⅰ)可知f(x)=2sin(2x+π6),∴函数f(x)的最小正周期T=2π2=π.(11分)函数f(x)的最大值为2.(13分)

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)=2√3sinxcosx-2sin²x+1 =√3sin2x+cos2x =2sin(2x+π/6) (2) 2kπ-π/2

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]

f(x)=2sinxcosx+2cos2x=sin2x+cos2x+1=2sin(2x+π4)+1,(Ⅰ)∵ω=2,∴函数f(x)的最小正周期为2π2=π,∵-1≤sin(2x+π4)≤1,∴1-2≤2sin(2x+π4)+1≤1+2,即1-2≤f(x)≤1+2,则函数f(x)的最小值为1-2;(Ⅱ)由α为锐角,f(α)=2,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

解:(Ⅰ)由f(x)=2 sinxcosx+2cos 2 x﹣1,得f(x)= (2sinxcosx)+(2cos 2 x)﹣1)= sin2x+cos2x=2sin(2x+ )所以函数f(x)的最小正周期为π.因为f(x)=2sin(2x+ )在区间[0, ]上为增函数,在区间[ , ]上为减函数,又f(0)=1,f(...

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