已知ab∈(3π/4,π),sin(a+b)=-3/5,sin(b-π/4)=12/13 求cos（a+π/4)值

题目

已知ab∈(3π/4,π),sin(a+b)=-3/5,sin(b-π/4)=12/13 求cos（a+π/4)值

答案

已知a,b∈(3π/4,π)
那么a+b∈(3π/2,2π),b-π/4∈(π/2,3π/4)
又sin(a+b)=-3/5,sin(b-π/4)=12/13
所以cos(a+b)=√[1-(-3/5)²]=4/5
cos(b-π/4)=-√[1-(12/13)²]=-5/13

所以cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
=(4/5)*(-5/13)+(-3/5)*(12/13)=-56/65