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哦～额……答案是sinα的平方

[sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简 化简sin(α-5π)/cos(3π-α)×cos(π/2-α)/sin(α-3π)×cos(8π-α)/sin(-α-4π) 化简[tan(2π-α)sin(-2π-α)cos(6π-α)]/[cos(α-π)sin(5π-α)] 化简tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α) 化简：[ cos(α-π/2)/sin(5π/2+α) ] ×sin(α-2π) × cos（2π-α）RT. 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 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+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) 2sinα*cosα*cos(2π-α)+cos(π+2α)*cosα*tanα化简! 1.化简根号2cos x - 根号6sin x2.已知sin(α-β)cosα-cos(β-α)sinα=3/5,β是第三象限角,求sin(β+5π/4)的值.3.化简sin(α-β)sin(β-λ)-cos(α-β)cos(λ-β) sin(α-β)sin(β-r)-cos(α-β)cos(r-β) 1.sin(α-β)sin(β-r)-cos(α-β)cos(r-β)2.（ tan4分之5π+tan12分之5π）/（1-tan12分之5π）3.[ sin（α+β）-2sinαcosβ]/2sinαsinβ+cos（α+β） 化简[sin(π/2-α）cos(π/2-α)]/cos(π+α)-[sin(π-α)cos(π/2-α)]/sin(π+α) 已知sin(3π+α)=2cos(α-4π)求cos(π/2 -α)+5sin(π/2 +α)/2cos(π+α)-sin(-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)] 化简：(sin²αtanα+cos²α/tanα+2sinαcosα)sinαcosα 已知α∈（π,2π） sinα+cosα=1/5 求sinα*cosα sinα-cosα 若sin(α-π)=2cos(2π-α),求（sinα+5cosα）/（sinα-3cosα）的值,