化简 :2sin(π+α)cos(π-α) 需具体过程,谢谢啦
化简cos(α-π/2)/sin(α+5π/2)*sin(α-2π)*cos(2π-α),因为我刚学,所以要具体点的过程
cos(π/2+α)=-sinα转换过程
化简[sin(π/2-α)cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π
α∈(0,π/2 ),比较 sin(cosα) 与cos(sinα)大小
已知(3sinα+cosα)/(3cosα-sinα)=2,则2-3sin(α-3π)sin(1.5π-α)-[cos(
化简:[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos(2π-α)
(sinα+cosα)^2化简
化简:{sin^2(α+π)*cos(α+π)*cos(-α-2π)}/{sin(-α-2π)*tan(π+α)*cos
化简:(1)sin(α+β)−2sinαcosβ2sinαsinβ+cos(α+β)
sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin(π-α)·cos(π/2+α)/sin(π+α)=
已知cos(π/2+α)=sin(α-π/2) 求sin(π-α)-cos(π+α)/cos(5π/2 -α)+2sin
已知α∈(π,2π) sinα+cosα=1/5 求sinα*cosα sinα-cosα