已知Sin(α+3π)=5/13,Cos(π/4-β)=4/5,-π/4
1个回答
展开全部
sin(a+3π/4)=5/13,其中,π/2<a+3π 4)="-12/13<br/" 4 cos(π/4-b)=4/5,其中,-π/2<π/4-b<0,则:sin(π/4-b)=-3/5
则:
cos(a-b)
=-cos[π+(a-b)]
=-cos[(a+3π/4)+(π/4-b)]
=-[cos(a+3π/4)cos(π/4-b)-sin(a+3π/4)sin(π/4-b)]
=-[(-12/13)×(4/5)-(5/13)×(-3/5)]
=33/65
则:
cos(a-b)
=-cos[π+(a-b)]
=-cos[(a+3π/4)+(π/4-b)]
=-[cos(a+3π/4)cos(π/4-b)-sin(a+3π/4)sin(π/4-b)]
=-[(-12/13)×(4/5)-(5/13)×(-3/5)]
=33/65
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