f(x)=sin(π-α)cos(2π-α)tan(-α+π)/-tan(α-π)sin(-π-α) 化简这个式子..

②若cos（已知α是第三象限角f(α)=sin(π-α)cos(2π-α)tan(-α-π)/tan（-α）sin（-π 化简：[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan( 求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan( 已知α为第三象限角,且f(α)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)tan(2π-α) 已知f(α)=sin(π-α)cos(2π-α)tan(π-α)/-tan(-π-α)sin(-π-α),若α是第三象限 已知f(α)=sin（π-α）cos(2π-α)tan(-α+3π/2)tan(-α-π) ／sin(-π-α).(1) 若f(α)=[sin(π-α）cos(2π-α）tan(-α+π）]/[-tan(-α-π）sin（-π-α）]化简 有关三角函数的题已知f(α)=sin(π-α)cos(2π-α)tan(-α+π)/-tan(-α-π)sin(-π-α α是第三象限角,f(α)={sin（α-π/2）cos（3π/2+α）tan（π-α）}/{tan（-α-π）sin（- 已知α是第三象限角,且f(α)=[sin(π+α)cos(2π-α)tan(-α+3π/2)tan(-α-π)]/sin 高一数学急!已知f(a)=sin^2(π-α)·cos(2π-α)·tan(-π+α)/sin(-π+α)·tan(-α 已知α是第三象限角,f(α)=[sin(π-a)cos(2π-a)tan(1.5π-a)tan(-a-π)]/sin(-