已知锐角α满足cosα-sinα=√5/5,则(sin2α-cos2α+1)/(1-tanα)=
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解决时间 2021-01-29 15:04
- 提问者网友:富士山上尢
- 2021-01-29 05:22
已知锐角α满足cosα-sinα=√5/5,则(sin2α-cos2α+1)/(1-tanα)=
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- 五星知识达人网友:怙棘
- 2021-01-29 06:58
cosα-sinα=√5/51-2sinacosa=1/52sinacosa=4/51+2sinacosa=9/5cosa+sina=3√5/5 (a为锐角,负值已舍)cosa=2√5/5sina=√5/5tana=1/2sin2a=2sinacosa=2 ×2√5/5× √5/5=4/5 cos2a= cos²a-sin²a=4/5-1/5=3/5分别代入(4/5-3/5+1)/(1-1/2)=6/5/1/2=12/5======以下答案可供参考======供参考答案1:cosα-sinα=√5/5 ,cosα=√5/5+sinα , cos^2α=1/5+2√5/5sinα+sin^2α=1-sin^2α2sin^2α+2√5/5sinα-4/5=0 , 2(sinα+√5/10)^2-9/10=0 ,sinα=-√5/10±3√5/10 ,sinα=√5/5 cosα=2√5/5(sin2α-cos2α+1)/(1-tanα)= [(cosα+sinα)^2-cos^2α+sin^2α]/(1-sinα/cosα) =(9/5-4/5+1/5)/(1-1/2) =12/5供参考答案2:cosα-sinα=√5/51-2cosαsinα=1/5sin2α=4/5cos2α=3/5cos2α=1-2sin²α =2cos²α-1tan²α=sin²α/cos²α =(1-cos2α)/(1+cos2α) =(2/5)/(8/5) =1/4tanα=1/2(sin2α-cos2α+1)/(1-tanα)= (4/5-3/5+1)/(1-1/2) =(6/5)/(1/2) =12/5
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- 1楼网友:迟山
- 2021-01-29 07:52
正好我需要
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