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)

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 03:54:56

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)
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)

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)
(1)-sina*(-cosa)+cosa*(-cosa)=sinacosa-(cosa)^2
(2)sinx*cosx/(-tanx)*(1/tanx)*(1/tanx)*(-sinx)=(cosx)^4/sinx

①[4sin(α-π)-sin(3π/2-α)]/[3cos(α-π/2)-5cos(α-5π)] ②sin^2-2sin^2αcosα-cos^2α 化简:sin(2π-α)cos(π+α)cos(11π/2-α)/cos(π-α)sin(3π-α)sin(-π-α)sin(9π/2+α) 化简sin(-π/2-α)sin(πα)cos(-α-π)/cos(π-α)sin(3π α) (1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 (3/2*π 化简:2sinα^2cosα-sin(3/2π-α)cos2α= 求值:(tan10°-√3)×cos10°/sin50° 求证:[sin(2α+β)]/sinα- 2cos(α+β)=sinβ/cosα 化简cosθ+cos(θ+2π/3)+cos(θ+4π/3) 证明:sin(α+β)sin(α-β)=sin^2α-sin^2β 设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π+α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2(-sinα)(-cosα)+cosα/[1+sin² ①sin(α+180°)cos(-α)sin(-α-180°)和②sin³(-α)cos(2π+α)tan(-α-π)化简 已知6sin²α+sinαcosα-2cos²α=0,α∈(π/2,π),求值1).(sinα-3cosα)/(sinα-cosα);2).sinαcosα-sin²α;3).sin²α-3cosαsinα-2 化简sin(α-5π)/cos(3π-α)×cos(π/2-α)/sin(α-3π)×cos(8π-α)/sin(-α-4π) 化简[sin(π/2-α)cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π+α) 【三角函数】已知sin(α-π)=2cos(2π-α),求[sin(π-α)+5cos(2π-α)]/[3cos(π-α)-sin(-α)] 若sin(α-π)=2cos(2π-α),求(sinα+5cosα)/(sinα-3cosα)的值, 化简sin(α-2π)+sin(-a-3π)cos(α-3π)/cos(π-α)-cos(-π-α)cos(α-4π) 已知tan(3π+α)=2,求:1、(sinα+cosα)²;2、sinα-cosα/2sinα+cosα 已知tan(π-α)=2,求sin²α-2sinαcosα-cos²α/4cos²α-3sin²α+1 已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值 若(sinα+cosα)/(sinα-cosα)=2 则sin(α-5π)·sin(3π/2-α)=?