已知tan(π/4+α)=2,则sinα-cosα|sinα+cosα=
展开全部
tan(π/4+α)=2
2=(tanπ/4 + tanα) / ( 1- tanπ/4tanα)
2(1- tanα) = 1+ tanα
3tanα= 1
tanα = 1/3
(sinα-cosα)/(sinα+cosα)
= 1 - 2cosα/(sinα+cosα)
= 1- 2/(tanα + 1)
= 1 - 2/( 1/3 + 1 )
= 1 - 3/2
= -1/2
2=(tanπ/4 + tanα) / ( 1- tanπ/4tanα)
2(1- tanα) = 1+ tanα
3tanα= 1
tanα = 1/3
(sinα-cosα)/(sinα+cosα)
= 1 - 2cosα/(sinα+cosα)
= 1- 2/(tanα + 1)
= 1 - 2/( 1/3 + 1 )
= 1 - 3/2
= -1/2
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
