[sin(2兀—¤)cos(兀+¤)cos(兀/2+¤)] / [cos(兀-a)sin(3兀-¤)sin(-¤)]
=[(-sin¤)(-cos¤)(-sin¤)]/[(-cos¤)(sin¤)(-sin¤)]
=-1
化简![sin(2兀—¤)cos(兀+¤)cos(兀/2+¤)] / [cos(兀-a)sin(3兀-¤)sin(-¤)]
化简![sin(2兀—¤)cos(兀+¤)cos(兀/2+¤)] / [cos(兀-a)sin(3兀-¤)sin(-¤)]
数学人气:520 ℃时间:2020-02-05 11:46:14
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