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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

(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。

解:f(x)=cos2x-sin2x+2sin2x. =cos2x+sin2x. ∴f(x)=√2sin(2x+π/4). 函数f(x)的最小值正周期T=2π/=π. 当sin(2x+π/4)=1时,函数f(x)取得最大值f(x)max=√2,此时x=2kπ+π/8, k∈Z.

f(x)=cos2x+23sinxcosx-sin2x=(cos2x-sin2x)+3(2sinxcosx)=cos2x+3sin2x=2sin(2x+π6),(1)∵ω=2,∴T=2π2=π,又正弦函数的递增区间为[2kπ-π2,2kπ+π2](k∈Z),∴2kπ-π2≤2x+π6≤2kπ+π2(k∈Z),解得:kπ-π3≤x≤kπ+π6(k∈Z),则函数f(x)...

(Ⅰ)由题意y=f(x)=sin2x+sinx?cosx+cos2x=1?cos2x2+12sin2x+cos2x(2分)=12(cos2x+sin2x)+12=22(22cos2x+22sin2x)+12(4分)∴y=22sin(2x+π4)+12(5分)∴y=f(x)的最小正周期T=π.(6分)(Ⅱ)由(Ⅰ)得∴y=22sin(2x+π4)+12由x∈[0,π2]得...

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

(1)∵f(x)=cos2x+23sinxcosx?sin2x=cos2x+3sin2x=2(12cos2x+32sin2x)=2sin(2x+π6),∴f(x)的最小正周期T=2π2=π.(2)∵x∈[0,π2]∴2x+π6∈[π6,7π6],∴sin(2x+π6)∈[?12,1],∴2sin(2x+π6)∈[?1,2],∴f(x)max=2,f(x)min=-1.

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