#p#因为tanα2,所以sin2α=2sinαcosα=2sinαcosα/(sinα+cosα)=2tanα/(1+tanα)=4/5cos2α=cosα-
因为tanα2,所以sin2α=2sinαcosα=2sinαcosα/(sin²α+cos²α)=2tanα/(1+tan²α)=4/5cos2α=cos²α-sin²α=(cos²α-sin²α)/(cos²α+sin²α)=(1-tan²α)/(1+tan²α)=-3/5
最多设置5个标签!
{var%20f='http://v.t.sina.com.cn/share/share.php?appkey=1515056452',u=z||d.location,p=['&url=',e(u),'&title=',e(t||d.title),'&source=',e(r),'&sourceUrl=',e(l),'&content=',c||'gb2312','&pic=',e(p||'')].join('');function%20a(){if(!window.open([f,p].join(''),'mb',['toolbar=0,status=0,resizable=1,width=440,height=430,left=',(s.width-440)/2,',top=',(s.height-430)/2].join('')))u.href=[f,p].join('');};if(/Firefox/.test(navigator.userAgent))setTimeout(a,0);else%20a();})(screen,document,encodeURIComponent,'','','http://www.jufenglt.com/data/attach/logo/logo.png', '推荐 的问题《若tanα=2,求sin2α和cos2α和值》','http://jufenglt.com/q-9373.html','页面编码gb2312|utf-8默认gb2312'));){kind=link}
因为tanα2,所以
sin2α=2sinαcosα
=2sinαcosα/(sin²α+cos²α)
=2tanα/(1+tan²α)
=4/5
cos2α=cos²α-sin²α
=(cos²α-sin²α)/(cos²α+sin²α)
=(1-tan²α)/(1+tan²α)
=-3/5