高一数学~任意角的三角函数
已知3sinα=-cosα,求
①2倍sinα的平方+3倍cosα的平方/sinα的平方+sinαcosα
②1+sinαcosα
求证:
tanαsinα/tanα-sinα=1+cosα/sinα
人气:247 ℃ 时间:2020-03-18 12:28:39
解答
1.
3sina=-cosa
tana=-1/3
[2(sina)^2+3(cosa)^2]/[(sina)^2+sinacosa]
=[2(tana)^2+3]/[(tana)^2+tana]
=-29/2
2.
1+sinacosa
=[(sina)^2+(cosa)^2+sinacosa]/[(sina)^2+(cosa)^2]
=[(tana)^2+1+tana]/[(tana)^2+1]
=7/10
3.
sina=cosatana
左边上下同时除以tana
得到sinatana/(tana-sina)=sina/(1-cosa)
要证明sinatana/(tana-sina)=(1+cosa)/sina
只需要证明sina/(1-cosa)=(1+cosa)/sina
只需要证明(sina)^2=(1-cosa)(1+cosa)=1-(cosa)^2.*
*式显然成立,所以,原命题为真...
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