已知0＜α＜π/2＜β＜π,cos（β-π/4π）=1/3，sin（α+β）=4/5，sin2β=-7/9。cos(α+π/4)的值
1个回答
展开全部
cos(β-π/4)=1/3
cos2(β-π/4)=2[cos(β-π/4)]^2-1=-7/9
cos(2β-π/2)=-7/9
cos(π/2-2β)=-7/9
sin2β=-7/9
cos(β-π/4)=1/3, sin(α+β)=4/5, 0＜α＜π/2＜β＜π
sin(β-π/4)=2√2/3
cos(α+β)=-3/5
cos(α+π/4)=cos[(α+β)+(π/4-β)]
=cos(α+β)cos(π/4-β)-sin(α+β)sin(π/4-β)
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
=(8√2-3)/15
cos2(β-π/4)=2[cos(β-π/4)]^2-1=-7/9
cos(2β-π/2)=-7/9
cos(π/2-2β)=-7/9
sin2β=-7/9
cos(β-π/4)=1/3, sin(α+β)=4/5, 0＜α＜π/2＜β＜π
sin(β-π/4)=2√2/3
cos(α+β)=-3/5
cos(α+π/4)=cos[(α+β)+(π/4-β)]
=cos(α+β)cos(π/4-β)-sin(α+β)sin(π/4-β)
=cos(α+β)cos(β-π/4)+sin(α+β)sin(β-π/4)
=(8√2-3)/15
本回答由网友推荐
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
