化简:cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
化简{sin(π+α)^3cos(-α)cos(π+α)}/{tan(π+α)^3cos(-π-α)^2}
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
化简:tan(π-α)·sin²(α+π/2)·cos(2π-α)/cos³(-α-π)·tan(α
已知tan(π+α)=-1/3 tan(α+β)=[sin(π-2α)+4(cosα)^2]/[10(cosα)^2-s
化简sin³(﹣α)cos(2π+α)tan(2π-α)/sin(α-2π)cos(α-3π/2)-tan(π
化简sin^2(α+π)cos(π+α)cot(-α+2π)/tan(π-α)cos^3(-α-π)
求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan(
已知tanα=1/3,tan(π-β)=-1/2,sin(α+β)-2sinαcosβ除以2sinαsinβ+cos(α
已知tan(3π+α)=2,求:1、(sinα+cosα)²;2、sinα-cosα/2sinα+cosα
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
已知tan(α+π/4)=2,则(2sinα+cosα)/(3cosα-2sinα)