急,已知f(α)=sin(π/2-α)cos(2π-α)tan(-α+3π)/tan(π+α)sin(π/2+α)(1)化简f(α)(2)若α是第三象限的角,且cos(α-3π/2)=1/5,求f(α)的值(3)若α=-1860°,求f(α)的值

问题描述:

急,已知f(α)=sin(π/2-α)cos(2π-α)tan(-α+3π)/tan(π+α)sin(π/2+α)
(1)化简f(α)
(2)若α是第三象限的角,且cos(α-3π/2)=1/5,求f(α)的值
(3)若α=-1860°,求f(α)的值

f(α)=sin(π/2-α)cos(2π-α)tan(-α+3π)/tan(π+α)sin(π/2+α)
=cosα cosα -tanα / tanα cosα
=-cosα

cos(α-3π/2)=1/5
-sinα=1/5
sinα=-1/5
f(α)=-cosα=2√6/5

f(α)=-cos(-1860°)=-cos60°=-1/2

1f(α)=sin(π/2-α)cos(2π-α)tan(-α+3π)/tan(π+α)sin(π/2+α)=cosα*cosα(-tanα)/(tanαcosα)=-cosα2∵cos(α-3π/2)=1/5∴-sinα=1/5,sinα=-1/5∵α是第三象限的角∴cosα=-√(1-sin...