化简[tan(2π-α)sin(-2π-α)cos(6π-α)]/[cos(α-π)sin(5π-α)]
展开全部
原式=[(-tana)(-sina)cosa]/[(-cosa)sina]
=(sina)^2/(-sinacosa)
=-tana
=(sina)^2/(-sinacosa)
=-tana
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
展开全部
[tan(2π-α)sin(-2π-α)cos(6π-α)]/[cos(α-π)sin(5π-α)]
=[tan(2π-α)sin(4π-2π-α)cos(4π+2π-α)]/[cos(π-α)sin(4π+π-α)]
=[-tanα*sin(2π-α)cos(2π-α)]/[cos(π-α)sin(π-α)]
=[-tanα*(-sinα)cosα]/[-cosαsinα]
=-[tanα*sinαcosα]/[cosαsinα]
=-tanα
=[tan(2π-α)sin(4π-2π-α)cos(4π+2π-α)]/[cos(π-α)sin(4π+π-α)]
=[-tanα*sin(2π-α)cos(2π-α)]/[cos(π-α)sin(π-α)]
=[-tanα*(-sinα)cosα]/[-cosαsinα]
=-[tanα*sinαcosα]/[cosαsinα]
=-tanα
本回答被提问者和网友采纳
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