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求F(x)=Cos2x+sinx是否为周期函数?答案化成%sin2x...

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)=2cos2x2+sinx=cosx+1+sinx=2sin(x+π4)+1.∴函数f(x)的最小正周期是2π,由2kπ?π2≤x+π4≤2kπ+π2,解得2kπ?3π4≤x≤2kπ+π4(k∈Z).∴函数f(x)的单调递增区间为[2kπ?3π4,2kπ+π4](k∈Z).)(2)由(1)函数f(x)=2sin(x+π4)+1.∵x...

f(x)=(cosx/2-√3sinx/2)(cosx/2+√3sinx/2) =(cosx^2-3sin^2x)/4 =(1+cos2x-3+3cos2x)/4 T=2π/2=π

f(x)=2sinxcosx+cos2x =sin2x+cos2x =√2sin(2x+π/4) T=2π/2=π sin函数的单增区间为 -π/2+2kπ,π/2+2kπ k为整数 即 -π/2+2kπ ≤2x+π/4 ≤π/2+2kπ 即 -3π/8+kπ≤x≤π/8+kπ ∵x∈[0,π/3] ∴f(x)max时 即sin=1 即f(x)max=√2 f(x)min=f(π/3)=√2sin(11π/12)

(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)

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。

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

f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =sinx(sinx*√3/2+cosx*1/2) =sin²x*√3/2+sinxcosx*1/2 =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4-√3/4*cos2x+1/4*sin2x =√3/4-1/2*(√3/2*cos2x+1/2*sin2x) =√3/4-1/2*(sinπ/3cos2x+cosπ/3sin2x) =√3/4-1/2*...

f(x)=4sinxcos(x+π/3)+根号3 =4sinx(1/2cosx-√3/2sinx)+√3 =2sinxcosx-2√3sin²x+√3 =sin2x-2√3sin²x+√3(sin²x+cos²x) =sin2x+√3cos²x-√3sin²x =sin2x+√3(cos²x-sin²x) =sin2x+√3con2x =2sin(2x+π/3) ...

f(x)=4cos∧2x+4√3sinxcosx-2 =2cos2x+2√3sin2x =4sin(2x+π/6), 它的最小正周期=π,最大值=4,此时,2x+π/6=(2k+1/2)π,k∈Z,解得x=(k+1/6)π。 其增区间是[(k-1/3)π,(k+1/5)π],对称轴由2x+π/6=(k+1/2)确定,即x=(k/2+1/6)π。 变式1 f(x)=(5/2)si...

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