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已知函数F(x)=sin(2x+π3)+sin(2x%π3)+2Cos2x...

f(x)=2sin2xcosπ3+cos2x=sin2x+cos2x=2sin(2x+π4)…(4分)(1)函数f(x)的最小正周期为2π2=π…(6分)(2)由题意知g(x)=f(x+3π8)+2=2sin(2x+3π4+π4)+2=?2sin2x+2…(8分)∵0≤x≤π∴0≤2x≤2π由g(x)在[0,π]上单调递减∴0≤2x≤π2,或3π2≤2x...

(1)∵f(x)=sin2x?cosπ3+cos2x?sinπ3+sin2x?cosπ3-cos2x?sinπ3+cos2x=sin2x+cos2x=2sin(2x+π4),∴函数f(x)的最小正周期T=2π2=π.(2)∵函数f(x)在区间[?π4,π8]上是增函数,在区间[π8,π4]上是减函数,又f(-π4)=-1,f(π8)=2,f(π4...

(1)f(x)=12sin2x+32cos2x+12sin2x-32cos2x=sin2x+cos2x+1=2sin(2x+π4)+1,∵ω=2,∴T=π;(2)由π2+2kπ≤2x+π4≤3π2+2kπ,k∈Z得:π8+kπ≤x≤5π8+kπ,k∈Z,∴f(x)的单调减区间为[kπ+π8,kπ+5π8],k∈Z;(3)作出函数y=f(x)在[-π4,π4]上的图象...

(1)∵得f(x)=sin2xcosπ3+sinπ3cos2x+sin2xcosπ3?sinπ3cos2x+cos2x+a=sin2x+cos2x+a∴f(x)=2sin(2x+π4)+a,…(4分)由f(x)max=2+1得a=1.…(3分)(2)∵g(x)=f(x+38π)?2=?2sin2x?1,…(4分)令-π2+2kπ≤2x≤π2+2kπ,k∈Z∴?π4+kπ≤x≤π4+kπ,k∈Z∴...

f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1 =sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x =2sin2xcosπ/3+cos2x =sin2x+cos2x =√2*(√2/2*sin2x+√2/2*cos2x) =√2*(sin2xcosπ/4+cos2xsinπ/4) =√2*sin(2x+π/4) T=2π/2=π x∈[-π/4,π/4] 2x∈[...

(Ⅰ)f(x)=sin(2x+π3)+sin(2x?π3)+2cos2x-1=(12sin2x+32cos2x)+(12sin2x-32cos2x)+cos2x=sin2x+cos2x=2sin(2x+π4)…(3分)由于x∈R,得2x+π4∈R,故sin(2x+π4)∈[-1,1],∴函数的值域为y∈[-2,2]…(6分)(Ⅱ)令-π2+2kπ<2x+π4<π2+2kπ,...

(1)f(x)=sin(2x+π3)+2cos2(π4-x)=sin(2x+π3)+[1+cos(π2-2x)]=12sin2x+32cos2x+1+sin2x=32sin2x+32cos2x+1=3sin(2x+π6)+1∴f(x)的最小正周期T=2π2=π,令2x+π6=π2+kπ(k∈Z),得x=π6+12kπ(k∈Z)∴f(x)的对称轴方程为x=π6+12kπ(...

(1)函数f(x)=sin(2x-π3)+cos(2x-π6)+2cos2x-1=sin2xcosπ3-cos2xsinπ3+cos2xcosπ6+sin2xsinπ6+cos2x (3分)=sin2x+cos2x (4分)=2sin(2x+π4) (5分)所以函数f(x)的最小正周期T=2π2=π.(6分)(2)∵f(x)在区间[-π4,π4]上是增...

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