已知向量a =(cos (2x -兀/3),cosx sinx )b =(1.cos x -sinx ),函数f (x
已知向量a =(cos (2x -兀/3),cosx sinx )b =(1.cos x -sinx ),函数f (x )=ab 求函数f (x )的单调递增区间
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向量a =(cos (2x -兀/3),cosx+sinx )
b =(1.cos x -sinx ),
f(x)=a·b
=cos(2x-π/3)+(cosx+sinx)(cosx-sinx)
=cos2xcosπ/3+sin2xsinπ/3+(cosx)^2-(sinx)^2
=1/2cos2x+√3/2sin2x+cos2x
=√3/2sin2x+3/2cos2x
=√3(1/2sin2x+√3/2cos2x)
=√3sin(2x+π/6)
由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z
得kπ-π/3≤x≤kπ+π/6,k∈Z
f(x)递增区间为[kπ-π/3,kπ+π/6｝,k∈Z
b =(1.cos x -sinx ),
f(x)=a·b
=cos(2x-π/3)+(cosx+sinx)(cosx-sinx)
=cos2xcosπ/3+sin2xsinπ/3+(cosx)^2-(sinx)^2
=1/2cos2x+√3/2sin2x+cos2x
=√3/2sin2x+3/2cos2x
=√3(1/2sin2x+√3/2cos2x)
=√3sin(2x+π/6)
由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z
得kπ-π/3≤x≤kπ+π/6,k∈Z
f(x)递增区间为[kπ-π/3,kπ+π/6｝,k∈Z
1年前
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