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数学
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已知f(a)=[sin(兀/2—a)cos(2兀—a)tan(-a+3兀)]÷[tan(兀+a )sin(兀/2+a)].(1) 化简f(a).
(2).若a是第三象限角,且cos(a—3兀/2)=1/5,求f(a)的值.
人气:344 ℃ 时间:2020-02-05 00:32:46
解答
(1)f(a)=[sin(兀/2-a)cos(2兀-a)tan(-a+3兀)]÷[tan(兀+a )sin(兀/2+a)]
=cosacosatan(-a)/[tanacosa]=-cosa.
(2) cos(a-3兀/2)=1/5,
cos(a-3兀/2+2兀)=1/5,
cos(兀/2+a) =1/5,
-sina=1/5,
sina=-1/5.
a是第三象限角,所以cosa=-2√6/5.
∴f(a)=-cosa=2√6/5.
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