Question

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

Answer 1

1
f(α)= sin(π-α)cos(2π-α)cos(-α+3/2π)/cos(π/2-α)sin(-π-α)
=[sinαcosα(-sinα)]/[sinα(-sinα)]=cosα
2
cos(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=-√(1-1/25)=-2√5/5
∴f(α)=cosα=-2√5/5