sin{π-α}cos{2π-α}tan{-α+π}/sin{π+α}化简
化简:[sin(﹣α)cos(π-α)]∕[tan(π+α)sin﹙3π/2+α)]
tan(π-α)=2,求(sinα-2cosα)/(3sinα+cosα)的值?
已知tan(π-α)=-2/1,则sinαcosα-2sin^2α=
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(
化简:f(a)=tan(-α-π)sin(-α-π)分之sin(α-2分之π)cos(2分之3π+α)tan(π-α)
若tanα=-½,求2sin²α+sinα*cosα-cos²α
已知tanα=2,sinα+cosα<0,求[sin(2π-α)*sin(π+α)*cos(-π+α)]/[sin(3π
若tanα=-2 ,则(3sin²α+sinα cosα-2cos²α)/1+2sin²α
化简sin³(﹣α)cos(2π+α)tan(2π-α)/sin(α-2π)cos(α-3π/2)-tan(π
[sin(2π+α)·tan(-α)·cos(-α)]÷[sin(-α)·cos(2π+α)·tan(2π+α)化简
sin^2α*tanα+cos^2α*cotα-2sinα*cosα(π-α)