∵a⊥c
∴cosα(sinx+2sinα)+sinα(cosx+2cosα)=0
sinxcosα+2sinαcosα+sinαcosx+2sinαcosα=0
sin(x+α)+2sin2α=0
又∵cos
=a·b/|a||b|
=(cosαcosx+sinαsinx)/(1*1)
=cos(x-α)
=cos(π/3)
=1/2
由积化和差公式,
sin(x+α)cos(x-α)
=1/2*[sin(x+α+x-α)+sin(x+α-x+α)]
=1/2*(sin2x+sin2α)
=-2sin2α*1/2
∴sin2x=-3sin2α
sin2(α+π/3)=-3sin2α
sin2αcos(2π/3)+cos2αsin(2π/3)=-3sin2α
-1/2*sin2α+√3/2*cos2α=-3sin2α
5sin2α=√3cos2α
∴tan2α=√3/5
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