已知f(α)= sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α) (1)化简f(_数学解答

已知f(α)= sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α) (1)化简f(α) (2)若α...
已知f(α)= sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)
(1)化简f(α)
(2)若α为第二象限的角,且cos(α-3/2π)=1/5,求f(α)的值

人气：158 ℃　时间：2020-07-15 05:38:13

解答

1f(α)= sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)=[sinαcosα(-sinα)]/[sinα(-sinα)]=cosα2cos(α-3/2π)=1/5==>sinα=-1/5 （需改?cos(α-3/2π)=-sinα矛盾）∵α为第二象限的角∴cosα...