若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+cos(-270º-α)】=?
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若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos(540º-α)+若sin(180°+α)=根号10/10,则【sin(-α)+si
若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+cos(-270º-α)】=?
若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+
若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+cos(-270º-α)】=?
若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+cos(-270º-α)】=?
哪里不明白再问我.
若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+若sin(180°+α)=根号10/10,则【sin(-α)+sin(-90º-α)】/【cos (540º-α)+cos(-270º-α)】=?
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