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tanx=-1/2,sinx平方+3sinx*cosx-1=?
人气:225 ℃ 时间:2020-03-25 19:07:34
解答
sin²x+3sinxcosx-1
=(3sinxcosx-cos²x)/(sin²x+cos²x)
=(3tanx-1)/(tan²x+1)
=(3×(-1/2)-1)/[(-1/2)²+1]
=(-5/2)/(5/4)
=-2
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