Sin(α+π)×Cos(π+α)×Cos(α+2π)/tan(π+α)×cos³(-α-π)
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:(1-sinα+cosα)/(1+sinα+cosα)=tan(π/4-α/2)
求证:1-2sinαcosα/cos²α-sin²α=tan(π/4-α)
已知tan(α+π/4)=3,求cosα+2sinα/cosα-sinα的值
试证明:1+2sinαcosα/cos平方α-sin平方α=tan(π/4-α)
化简:{sin^2(α+π)*cos(α+π)*cos(-α-2π)}/{sin(-α-2π)*tan(π+α)*cos
化简sin³(﹣α)cos(2π+α)tan(2π-α)/sin(α-2π)cos(α-3π/2)-tan(π
已知tan(π-α)=2,求sin²α-2sinαcosα-cos²α/4cos²α-3s
①sin(α+180°)cos(-α)sin(-α-180°)和②sin³(-α)cos(2π+α)tan(-
求证:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]=
求证:[tan(2π-α)cos(3π/2-α)cos(6π-α)]/[sin(α+3π/2)cos(α+3π/2)]