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)
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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);
(2)若cos(α-3π/2)=1/5,求f(α); (3)若α=-1860°,求f(α);
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
再问: =sinacosacota/(-cosa) =-cosa (a≠π/2+kπ) 这一部能在详细一点么?怎么得出的-cosa ?
再答: cota=cosa/sina sinacosacota/(-cosa) =-sinacosacosa/sinacosa =-cosa
=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
再问: =sinacosacota/(-cosa) =-cosa (a≠π/2+kπ) 这一部能在详细一点么?怎么得出的-cosa ?
再答: cota=cosa/sina sinacosacota/(-cosa) =-sinacosacosa/sinacosa =-cosa
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