已知sin(α+π/6)=1/3,则cos(2α+π/3)=?sin(π/6-2α)=?

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已知sin(α+π/6)=1/3,则cos(2α+π/3)=?sin(π/6-2α)=?已知sin(α+π/6)=1/3,则cos(2α+π/3)=?sin(π/6-2α)=?已知sin(α+π/6)

已知sin(α+π/6)=1/3,则cos(2α+π/3)=?sin(π/6-2α)=?
已知sin(α+π/6)=1/3,则cos(2α+π/3)=?sin(π/6-2α)=?

已知sin(α+π/6)=1/3,则cos(2α+π/3)=?sin(π/6-2α)=?
cos(2α+π/3)=cos2*[α+π/6]=1-sin^2[α+π/6]=1-1/9=8/9
sin(π/6-2α)=cos[π/2-(π/6-2α)]=cos[2α+π/3]=8/9

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