2个回答
展开全部
cos(α-π/2) = cos(π/2-α) = sinα
sin(5π/2+α) = sin(π/2+α) = cosα
sin(α-π) =-sin(π-α) = -sinα
cos(2π-α) = cosα
[cos(α-π/2)/sin(5π/2+α) ] .sin(α-π).cos(2π-α)
=(sinα/cosα) .(-sinα)(cosα)
=-(sinα)^2
sin(5π/2+α) = sin(π/2+α) = cosα
sin(α-π) =-sin(π-α) = -sinα
cos(2π-α) = cosα
[cos(α-π/2)/sin(5π/2+α) ] .sin(α-π).cos(2π-α)
=(sinα/cosα) .(-sinα)(cosα)
=-(sinα)^2
本回答被网友采纳
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