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已知tanα=1/2,求(sinα)^2+3sinα·cosα的值
人气:280 ℃ 时间:2020-01-30 04:34:02
解答
(sinα)^2+3sinα·cosα=[(sinα)^2+3sinα·cosα ]÷1=[(sinα)^2+3sinα·cosα]÷((sinα)^2 +(cosα)^2) (1的代换)=[ (tanα)^2 +3tanα] ÷[(tanα)^2 +1] (分子分母同除以(cosα)^2 )=(0....
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