化简:(2)cos(70°+α)sin(170°+α)-sin(70°+α)cos(10°+α)

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 20:07:31
化简:(2)cos(70°+α)sin(170°+α)-sin(70°+α)cos(10°+α)

化简:(2)cos(70°+α)sin(170°+α)-sin(70°+α)cos(10°+α)
化简:(2)cos(70°+α)sin(170°+α)-sin(70°+α)cos(10°+α)

化简:(2)cos(70°+α)sin(170°+α)-sin(70°+α)cos(10°+α)
用a代替
错了,应该是sin(170-a)
原式=cos(70+a)sin[180-(170-a)]-sin(70+a)cos(10+a)
=cos(70+a)sin(10+a)-sin(70+a)cos(10+a)
=sin[(10+a)-(70+a)]
=sin(-60)
=-√3/2

(sinα+cosα)^2化简 sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β)化简 化简sin(α+β)-2sinαcosβ 除以 2sinαsinβ+cos(α+β) 化简(2sinα+cosα)²+(2cosα-sinα)² 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 化简:(2)cos(70°+α)sin(170°+α)-sin(70°+α)cos(10°+α) ①sin(α+180°)cos(-α)sin(-α-180°)和②sin³(-α)cos(2π+α)tan(-α-π)化简 求证 1+sinα+cosα+2sinαcosα/求证 (1+sinα+cosα+2sinαcosα)/(1+sinα+cosα)=sinα+cosα 化简 根号2cosα-根号2sinα化简(1) 根号2cosα-根号2sinα(2)sinα+cosα 化简[sin(π/2-α)cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π+α) 2sin²α-sinαcosα/sinαcosα+cos²α (tan=2) 已知sinα+3cosα=2,求(sinα-cosα)/(sinα+cosα)的值 已知sinα+3cosα=2,求(sinα-cosα)/(sinα+cosα)的值 化简(1-sinα+cosα/1-sinα-cosα )+(1-sinα-cosα/1-sinα+cosα) 化简:sin²(α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π)1.cos(27°+α)cos(33°-α)-sin(27°+α)sin(33°-α) 2.sin²(α+π)cos(-α+π)/tan(α+π)tan(α+2π)cos²(-α-π) 求证:2(cosα-sinα)/(1+sinα+cosα) = cosα/(1+sinα)- sinα/(1+cosα) 证明2(cosα-sinα)/(1+sinα+cosα)=cosα/(1+sinα)-sinα/(1+cosα) 化简sin^4x-sin^2+cos^2化简(1)sin^4x-sin^2+cos^2(2)根号(1+cosα/1-cosα)+根号(1-cosα/1+cosα)(180°<α<270°)