若sin(α-360°)-cos(180°-α)=m,则sin(180°+α)·cos(180°-α)等于

Cos(180°+α)*sin（α+360°)/sin(-α-180°)*cos(-180°-α) 若sin(α-360°)-cos(180°-α)=m,则sin(180°+α)·cos(180°-α)等于 化简cos(540°-α)sin(α-360°)/sin(-α+180°)cos(180°+α) .[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α 1.sin的三次方(-α)cos(2π+α)tan(-α-π)2.cos(180°+α)sin(α+360°)/sin(-α-180°)cos(-180°-α)3.sin(2π-α)cos(π+α)cos(π/2+α)sin(9π/2+α)/cos(π-α)sin(3π-α)sin(-π-α)4.已知sinα=-3/5,求cosα,tanα的值5.已知tanα=-根号下3,求 cos(540-α)sin(α-360)/sin(-α+180)cos(180+α) 若sinα*cosα 若sinαcosα 若sinαcosα 若sinαcosα 已知sin（360°+α）-cos（180°-α）=m则sin（180°+α）cos（180°-α）=? 求证：(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3 化简,cos²(-α)-tan(360°+α)/sin(-α) 化简cos^2(-α)-[tan(360°+α)/sin(-α)] sin(a-360°)-cos(180°-a)= 急~一道高一数学题 若sin(180°+α)+cos(90°+α)=-α,则cos（270°-α）+2sin(360°-α）的值是要过程 谢谢 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β=α 30°,则sin²α cos²β sinαcosβ等于若β=α+30°,则sin²α+cos²β+sinαcosβ等于