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

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∈[...

供参考。

f(x)=2cos²x-2sin(x+ 3π/2)cos(x- π/3) -3/2 =2cos²x+2cosx(cosxcosπ/3 +sinxsinπ/3) -3/2 =2cos²x+2cosx[½cosx+(√3/2)sinx] -3/2 =2cos²x+cos²x+√3sinxcosx -3/2 =3cos²x+√3sinxcosx -3/2 =(3/2)(2cos²...

f(x)=sin(2x-π6)+2cos2x-1=32sin2x-12cos2x+cos2x=32sin2x+12cos2x=sin(2x+π6)由-π2+2kπ≤2x+π6≤π2+2kπ(k∈Z)得:-π3+kπ≤x≤π6+kπ(k∈Z)由π2+2kπ≤2x+π6≤3π2+2kπ(k∈Z)得:π6+kπ≤x≤2π3+kπ(k∈Z)故f(x)的单调递增区间是[-π3+kπ,π6+kπ]...

sin(π/3-x)=cos[π/2-(π/3-x)]=cos(x+π/6) 这样再利用降次公式应该就好算了。 f(x)=cos(2x-π/3)+sin方(π/6+x)+cos方(x+π/6) =cos(2x-π/3)+1/2-1/2cos(π/3+2x)+1/2+1/2cos(π/3+2x) =cos(2x-π/3)+1 于是T=π, 对称轴x=2x-π/3=kπ(k属于Z) x=π/6+kπ...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

f(x)=sin2x+3cos2x=2(12sin2x+32cos2x)=2sin(2x+π3).∵f(x)的图象关于点(x0,0)成中心对称,∴f(x0)=0,即2sin(2x0+π3)=0,∴2x0+π3=kπ,x0=kπ2?π6,k∈Z,∵x0∈[0,π2],∴x0=π3.故选:C.

解:先用降幂公式把函数化为:f(x)=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1 (1)最小值为-2,最小正周期为π (2)由f(C)=0知sin(2C-π/6)=1,从而可得C=π/3,再由余弦定理知:c^2=a^2+b^2-2abcosC 3=a^2+4a^2-2a*2acosπ/3,解得a=1,故b=2

函数f(x)=3sin(ωx-π6)(ω>0)的对称轴方程为ωx-π6=kπ+π2,即 x=kπω+2π3ω,k∈z.g(x)=2cos(2x+φ)(0<φ<π)的图象的对称轴为 2x+φ=kπ,即 x=kπ2-φ2,k∈z.函数f(x)=3sin(ωx-π6)(ω>0)和g(x)=2cos(2x+φ)(0<φ<π)的图象的对...

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