[tan(2π-α)·sin(2π-α)·cos(6π-α)·sin(3π/2 -α)]/[cos(-α)·sin(5π
化简:[sin(﹣α)cos(π-α)]∕[tan(π+α)sin﹙3π/2+α)]
已知tan(π-α)=-2/1,则sinαcosα-2sin^2α=
若(sinα+cosα)/(sinα-cosα)=2 则sin(α-5π)·sin(3π/2-α)=?
tan(π-α)=2,求(sinα-2cosα)/(3sinα+cosα)的值?
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
已知tanα=3,求2sinα+cosα/5sinα-3cosα的值
已知tanα=2,sinα+cosα<0,求[sin(2π-α)*sin(π+α)*cos(-π+α)]/[sin(3π
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
求证:[sin^2(π+α)*cos(π/2-α)+tan(2π-α)*cos(-α)]/-sin^2(-α)+tan(
已知sin a=2cos a 计算⑴(sinα+2cosα)/(5cosα-sinα) ⑵tan(α+π/4)
已知tan(α+π/4)=2,则(2sinα+cosα)/(3cosα-2sinα)