已知sin(α+π/3)=√2/4,π/6<α<2π/3,则cos(2π/3 + α)=?
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已知sin(α+π/3)=√2/4,π/6<α<2π/3,则cos(2π/3 + α)=?
已知sin(α+π/3)=√2/4,π/6<α<2π/3,则cos(2π/3 + α)=?
已知sin(α+π/3)=√2/4,π/6<α<2π/3,则cos(2π/3 + α)=?
∵π/6<α<2π/3
∴ π/2<α+π/3<π
∴ cos(α+π/3)
∵sin(α+π/3)+sinα=-(4√3)/5
∴sinαcosπ/3+cosαsinπ/3+sinα=-4√3/5
∴3/2*sinα+√3/2*cosα=-4√3/5
∴√3/2sinα+1/2*cosα=-4/5
∴sin(α+π/6)=-4/5
∵-π/2<α<0
∴-π/3<α+π/6<π/6
∴cos(α+π/6)=3/5
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∵sin(α+π/3)+sinα=-(4√3)/5
∴sinαcosπ/3+cosαsinπ/3+sinα=-4√3/5
∴3/2*sinα+√3/2*cosα=-4√3/5
∴√3/2sinα+1/2*cosα=-4/5
∴sin(α+π/6)=-4/5
∵-π/2<α<0
∴-π/3<α+π/6<π/6
∴cos(α+π/6)=3/5
∴cosα=cos[(α+π/6)-π/6]
=cos(α+π/6)cosπ/6+sin(α+π/6)sinπ/6
=3/5*√3/3-4/5*1/2
=(3√3-4)/6望采纳,谢谢
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