已知角a终边经过点P(x,-√2)(x≠0),且cos a=√3x/6 求sin a 、tan a
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人气:368 ℃ 时间:2020-05-13 19:11:01
解答
P(x,-√2)那么斜边为√(x^2+(√2)^2)=√(x^2+2)cosa=x/√(x^2+2)=√3x/6两边平方x^2/(x^2+2)=3x^2/361/(x^2+2)=1/12x^2=10x=√10所以sina=-√2/√((√10)^2+2)=-√6/6tana=-√2/√10=-√5/5...其实我已经弄懂了而且你的回答是错的这道题应该分类讨论 还有x=-根号10的情况
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