sin(α+π/4)=1/3,则sin2α
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sin(α+π/4)=1/3,则sin2αsin(α+π/4)=1/3,则sin2αsin(α+π/4)=1/3,则sin2αsin(α+π/4)=sinα*(√2/2)+cosα*(√2/2)=1/
sin(α+π/4)=1/3,则sin2α
sin(α+π/4)=1/3,则sin2α
sin(α+π/4)=1/3,则sin2α
sin(α+π/4)=sinα*(√2/2)+cosα*(√2/2)=1/3
∴sinα+cosα=√2/3
∴sin2α=2sinαcosα=(sinα+cosα)^2-1=-7/9
sin2α=sin[2(α+π/4)-π/2]=-COS2(α+π/4)=-(1-2(sin(α+π/4))^2)=-7/9
(sin(α+π/4))^2=1/9
(1-cos(2α+π/2))/2=1/9
cos(2α+π/2)=7/9
-sin(2α+π/2)=7/9
sin(2α)=-7/9
答案见图片
sin(α+π/4)=1/3,则sin2α
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