化简 ：2sin(π+α）cos（π-α） 需具体过程,谢谢啦

化简[sin(π/2-α）cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π+α) 化简：2sin（π+α）cos（π-α） 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：[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos（2π-α）RT. α∈（0,π／2 ）,比较 sin(cosα) 与cos(sinα)大小 化简［sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)］/［cos(π-α)sin(π+α)sin(-π-α)sin［（...化简［sin(2π-α)cos(π+α)cos(π/2+α)cos(11π/2-α)］/［cos(π-α)sin(π+α)sin(-π-α)sin［（9π/2）+α］］ [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简 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 sin(α-π)=2cos（α-2π）求cos平方α-sin平方αcosα 设α∈（0,π/2）,则cosα,sin(cosα),cos(sinα)的大小关系为什么? 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 若α∈（0,π）,化简[1+sinα+cosα）(sinα/2-cosα/2)]/根号（2+2cosα） 2sinα*cosα*cos(2π-α)+cos(π+2α)*cosα*tanα化简! ①sin(α+180°)cos(-α)sin(-α-180°)和②sin³（-α）cos（2π+α）tan（-α-π）化简 若cosα>sinα,(-π/2