(sinα+cosα)/(sinα-cosα)+cos^2α的值
(2)若f(α)=3/5,求sinα,tanα的值
已知f(α)=[sin(π-α)*cos(2π-α)*tan(-α+3π/2)]/cos(-π-α)(1)若2f(π+α)=f(π/2+α),求
答案:2 悬赏:60
解决时间 2021-11-07 05:11
- 提问者网友:柠檬香
- 2021-11-06 10:50
最佳答案
- 二级知识专家网友:劳资的心禁止访问
- 2021-11-06 12:24
解:f(α)=sin(π-α)*cos(2π-α)*tan(-α+3π/2)/cos(-α-π)
=-sin(α)*cos(α)*cot(α)/cos(α)
=-sin(α)*cot(α)
=-cosα.
(1) ∵ 2f(π+α)=f(π/2+α)
∴ -2cos(π+α)=-cos(π/2+α)
∴2cosα=sinα
∴tanα=2
(sinα+cosα)/(sinα-cosα)+cos^2α
=(tanα+1)/(tanα-1)+1/((tanα)^2+1)
=3+1/5=16/5.
(2)f(α)=3/5,即-cosα=3/5, ∴cosa=-3/5,sina=4/5或-4/5,tana=4/3或-4/3.
如果帮到你,请记得采纳,O(∩_∩)O谢谢
=-sin(α)*cos(α)*cot(α)/cos(α)
=-sin(α)*cot(α)
=-cosα.
(1) ∵ 2f(π+α)=f(π/2+α)
∴ -2cos(π+α)=-cos(π/2+α)
∴2cosα=sinα
∴tanα=2
(sinα+cosα)/(sinα-cosα)+cos^2α
=(tanα+1)/(tanα-1)+1/((tanα)^2+1)
=3+1/5=16/5.
(2)f(α)=3/5,即-cosα=3/5, ∴cosa=-3/5,sina=4/5或-4/5,tana=4/3或-4/3.
如果帮到你,请记得采纳,O(∩_∩)O谢谢
全部回答
- 1楼网友:啵啵桃汀
- 2021-11-06 13:20
(1) f(α)=[sinα·﹙-cosα﹚·cotα·(-tanα)]/sinα=cosα
(2) f(-31π/3)=cos(-31π/3)=cos(-π/3)=cos(π/3)=1/2
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