{[sin(π-α)-cosα]^2}/{cot[(3π/2)+α]} - 2cos(α+5π/6)*cos(α-5π/6)

若cosα>sinα,(-π/2 sinα-2cosα分之sinα+cosα 3sinα=2cosα,则 cosα-sinα/cosα+sinα/ cos3sinα=2cosα,则cosα-sinα/cosα+sinα/cosα+sinα/cosα-sinα sin(α+β)-2cosαsinβ/2cosαcosβ-cos(α+β) 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 α∈（0,π／2 ）,比较 sin(cosα) 与cos(sinα)大小 比较大小sin(cosα)与cos(sinα)（0＜α＜π/2) sin(α+β)-2sinαcosβ/2sinαsinβ+cos(α+β) 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) (1-cos-sin)(1-sin+cos)/sin^2α-sinα cos^2(π/4+α)=(cos(π/4)cosα-sin(π/4)sinα)² cos^2(π/4+α)是怎么等于(cos(π/4)cosα-sin(π/4)sinα)²的? [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简 sin(α-π)=2cos（α-2π）求cos平方α-sin平方αcosα (1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 （3/2*π 化简sin(-π/2-α)sin(πα)cos(-α-π)/cos(π-α)sin(3π α) 求值【sin(2π-α)sin(π+α)cos(-π-α)】/【sin(3π-α)-cos(π-α)】 设α∈（0,π/2）,则cosα,sin(cosα),cos(sinα)的大小关系为什么? cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα