sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)急!

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/18 21:32:47
sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)急!sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)急!sin(54

sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)急!
sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)
急!

sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)急!
sin(540°+α)cos(360°-α)/[sin(450°+α)tan(900°-α)]
=sin(180°+α)cosα/[sin(90°+α)tan(180°-α)]
=-sinαcosα/[-cosα.tanα]
=-sinαcosα/[-sinα]
=cosα

化简cos(540°-α)sin(α-360°)/sin(-α+180°)cos(180°+α) 化简sin(540°+α)cos(360°-α)/sin(450°+α)cos (900°-α) .[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α) 化简,cos²(-α)-tan(360°+α)/sin(-α) 化简cos^2(-α)-[tan(360°+α)/sin(-α)] 求证:(1/sinα-sin(180°+α))/(1/cos(540°-α)+cos(360°-α))=1/(tanα)^3 sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)急! sin(α+30°)-sin(α-30°)/cosα 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) cos(α+30°)cosα+sin(α+30°)sinα等于? sin(α+30°)cosα+cos(α+30°)sinα= 为什么sin(90°+α)=cosα ,cos(90°+α)=-sinα 利用和差角公式化简 (2)sin(π/3+α)+sin(π/3-α)(2)sin(π/3+α)+sin(π/3-α)(3)cos(π/4+α)-cos(π/4-α)(4)cos(60°+α)+cos(60°-α)(5)sin(α-β)cosβ+cos(α-β)sinβ(6)cos(α+β)cosβ+sin(α+β)sinβ 已知cos(α+β) =cosαsinβ-sinαsinβ,求sin15°=急阿急阿! 求证下列各恒等式:1.sin(30°+α)+cos(60°+α)=cosα1.sin(30°+α)+cos(60°+α)=cosα2.sin(α+β)sin(α-β)=sin²α-sin²β3.cos(α+β)cos(α-β)=cos²β-sin²α 化简①sin(π+α)×cos(3π/α)+sin(π/2+α)×cos(π+α)②sin(540°-x)×cos(x-360°)/tan(900°-x)×tan(450°-x)×tan(810°-x)×sin(-x)