急,已知f(α)=sin(π/2-α)cos(2π-α)tan(-α+3π)/tan(π+α)sin(π/2+α)

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
1个回答 分类:数学 2014-12-06

问题解答:

我来补答
1
f(α)=sin(π/2-α)cos(2π-α)tan(-α+3π)/tan(π+α)sin(π/2+α)
=cosα*cosα(-tanα)/(tanαcosα)
=-cosα
2
∵cos(α-3π/2)=1/5
∴-sinα=1/5,sinα=-1/5
∵α是第三象限的角
∴cosα=-√(1-sin²α)=-2√6/5
∴f(α)=-cosα=2√6/5
3
f(α)=-cosα=-cos(-1860º)
=-cos1860º
=-cos(5*360º+60º)
=-cos60º=-1/2
 
 
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