已知f(α)=sin(π−α)•cos(2π−α)•tan(−π−α)sin(−π−α)
已知f(α)=sin(π−α)•cos(2π−α)•tan(−π−α)sin(−π−α)
已知tanα=2,求sinα−3cosαsinα+cosα
已知α为第二象限角,f(α)=sin(α−π2)cos(3π2+α)tan(π−α)tan(−α−π)sin(−α−π)
已知tanα=3,求2cos(π−α)−3sin(π+α)4cos(−α)+sin(2π−α)
证明:1+sinα−cosα1+sinα+cosα=tanα2
已知α是第二象限角,f(α)=sin(π−α)tan(−α−π)sin(π+α)cos(2π−α)tan(−α).
已知sin(3π−α)=−2sin(π2+α),则sinαcosα=( )
若tanα=2,求2sinα+cosαsinα−cosα
已知α为锐角,tanα=3,求sinα−cosαsinα+2cosα
若sinα+cosαsinα−cosα=2,则sin(α-5π)•sin(3π2-α)等于( )
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
若sinα+cosα2sinα−cosα=2,则tanα=( )