已知向量a=(-√3sinωx,cosωx),向量b=(cosωx,cosωx)(ω>0),令函数f(x)=向量a向量b,
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f(x)=(-√3sinωx,cosωx)(cosωx,cosωx)
=-√3sinωxcosωx+cos²ωx
=-√3/2sin2ωx+(cos2ωx+1)/2
=cos2ωxcosπ/3-sin2ωxsinπ/3+1/2
=cos(2ωx+π/3)+1/2
2ω=2π/T=2
ω=1
∴f(x)=cos(2x+π/3)+1/2
令2kπ≤2x+π/3≤π+2kπ
得-π/6+kπ≤x≤π/3+kπ
令-π+2kπ≤2x+π/3≤2kπ
得-2π/3+kπ≤x≤-π/6+kπ
∴f(x)于[-2π/3+kπ,-π/6+kπ]↗
于[-π/6+kπ,π/3+kπ]↘(k∈Z)
=-√3sinωxcosωx+cos²ωx
=-√3/2sin2ωx+(cos2ωx+1)/2
=cos2ωxcosπ/3-sin2ωxsinπ/3+1/2
=cos(2ωx+π/3)+1/2
2ω=2π/T=2
ω=1
∴f(x)=cos(2x+π/3)+1/2
令2kπ≤2x+π/3≤π+2kπ
得-π/6+kπ≤x≤π/3+kπ
令-π+2kπ≤2x+π/3≤2kπ
得-2π/3+kπ≤x≤-π/6+kπ
∴f(x)于[-2π/3+kπ,-π/6+kπ]↗
于[-π/6+kπ,π/3+kπ]↘(k∈Z)
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