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tanα=-2,则sin平方α+sinαcosα=?
人气:171 ℃ 时间:2020-05-02 09:42:02
解答

(sina)²+sinacosa
=sin²a+sinacosa
=(sin²a+sinacosa)/(sin²a+cos²a)
=(tan²a+tana)/(tan²a+1)
=(2²+2)/(2²+1)
=6/5
=5分之6
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