若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)
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若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)若sin(180°+α)=-√10/10,α∈(0°
若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)
若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)
若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)
sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-α)式子有误.是不是:
[sin(-α)+(sin-90°-α)]/[cos(540°-α)+cos(-270°-α)]
若sin(180°+α)=-√10/10,α∈(0°,90°),求sin(-α)+(sin-90°-α)/cos(540°-α)+(-270°-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)】=?
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