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f(α)=cot(-α-π)sin(-π-α)分之(sin(π-α)cos(2π+α)tan(-α+3π/2))?(1)化简f(α)(2)若α为第三象限角且cos(α-2分之3π)=5分之1.求f(α)的值?
题目详情
f(α)=cot(-α-π)sin(-π-α)分之(sin(π-α)cos(2π+α)tan(-α+3π/2))?
(1)化简f(α) (2)若α为第三象限角且cos(α-2分之3π)=5分之1 .求f(α)的值?
(1)化简f(α) (2)若α为第三象限角且cos(α-2分之3π)=5分之1 .求f(α)的值?
▼优质解答
答案和解析
f(α)=[sin(π-α)cos(2π+α)tan(-α+3π/2)]/cot(-α-π)sin(-π-α)
=sinαcosαcotα/(-cotαsinα)=-cosα
cos(α-3π/2)=1/5=-sinα,sinα=-1/5,cosα=-2√6/5 ,f(α)=-cosα=-2√6/5
=sinαcosαcotα/(-cotαsinα)=-cosα
cos(α-3π/2)=1/5=-sinα,sinα=-1/5,cosα=-2√6/5 ,f(α)=-cosα=-2√6/5
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