sin(-α-5π)cos(α-π/2)-tan(α-3π/2)*(2π-α)

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sin(-α-5π)cos(α-π/2)-tan(α-3π/2)*(2π-α)

sin(-α-5π)cos(α-π/2)-tan(α-3π/2)*(2π-α)
sin(-α-5π)cos(α-π/2)-tan(α-3π/2)*(2π-α)

sin(-α-5π)cos(α-π/2)-tan(α-3π/2)*(2π-α)
(α-3π/2)*(2π-α)
这是连在一起的吗?

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 [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简 若cosα>sinα,(-π/2 化简sin(α-5π)/cos(3π-α)×cos(π/2-α)/sin(α-3π)×cos(8π-α)/sin(-α-4π) 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) 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)] 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, 已知sin a=2cos a 计算⑴(sinα+2cosα)/(5cosα-sinα) ⑵tan(α+π/4) 若(sinα+cosα)/(sinα-cosα)=2 则sin(α-5π)·sin(3π/2-α)=? 若sinα+cosα/sinα-cosα=2,则sin(α-5π)*sin(3π/2-α)等于? 若(sinα+cosα)/(sinα-cosα)=2,求sin(α-5π)sin(3π/2-α)的值 sinα+cosα=1/5,且π/2 sin(π+α)+cos(5π-α))/(sin(α-6π)+2cos(2π+α)=(-sinα-cosα)/(sinα+2cosα)为什么相等 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) ①[4sin(α-π)-sin(3π/2-α)]/[3cos(α-π/2)-5cos(α-5π)] ②sin^2-2sin^2αcosα-cos^2α