已知α是第三象限角,且f(α)=sin（π-α)cos(2π-α)tan(-α+3π/2）/cos(-α -π) (1）化简f(α);
问题描述：
已知α是第三象限角,且f(α)=sin（π-α)cos(2π-α)tan(-α+3π/2）/cos(-α -π) (1）化简f(α);
(2)若cos(α-3π/2)=1/5,求f(α); (3)若α=-1860°,求f(α);
答
f(α)=sin（π-α)cos(2π-α)tan(-α+3π/2）/cos(-α -π)
=sinacosacota/(-cosa)
=-cosa (a≠π/2+kπ)
cos(α-3π/2)=1/5
sina=1/5
所以cosa=±2√6/5
所以f(α)=-cosa=±2√6/5
α=-1860°
所以f(α)=-cosa=-cos(-1860)=-cos(1800-1860)=-cos(-60)=-1/2=sinacosacota/(-cosa) =-cosa(a≠π/2+kπ)这一部能在详细一点么？怎么得出的-cosa？cota=cosa/sinasinacosacota/(-cosa)=-sinacosacosa/sinacosa=-cosa