已知sin(360°+α)-cos(180°-α)=m则sin(180°+α)cos(180°-α)=?
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已知sin(360°+α)-cos(180°-α)=m则sin(180°+α)cos(180°-α)=?
已知sin(360°+α)-cos(180°-α)=m
则sin(180°+α)cos(180°-α)=?
已知sin(360°+α)-cos(180°-α)=m则sin(180°+α)cos(180°-α)=?
sin(360°+α)-cos(180°-α)= sinα +cosα =m
(sinα +cosα)^2 = 1+2sinα*cosα =m^2,sinα*cosα=(m^2-1)/2
sin(180°+α)cos(180°-α)= -sinα(-cosα)=sinα*cosα=(m^2-1)/2
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已知sin(360°+α)-cos(180°-α)=m则sin(180°+α)cos(180°-α)=?
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