Question

已知π /4<a<3π /4,0<b<π/4,cos(π/4+a)=-3/5,sin(3π/4+b)=5/13,求sin（a+b）

Answer 1

cos(π/4+a)=-3/5
sin(π/4+a)=4/5
sin(3π/4+b)=5/13
cos(3π/4+b)=-12/13
sin(π/4+a+3π/4+b)
=sin(a+b+∏)
=-sin(a+b)
=sin(π/4+a)cos(3π/4+b)+cos(π/4+a)sin(3π/4+b)
=(4/5)(-12/13)+(-3/5)(5/13)
=-63/65
sin（a+b）=63/65

Answer 2

π /4<a<3π /4,0<b<π/4
π/2<π/4+a<π 3π/4<3π/4+b<π
π/2<π/4+b<π
cos(π/4+a)=-3/5, sin(3π/4+b)=cos(π/4+b)=5/13
sin(π/4+a)=4/5 sin(π/4+b)=12/13
sin(a+b)=-cos(π/2+a+b)=-cos[(π/4+a)+(π/4+b)]=sin(π/4+a)sin(π/4+b)-cos(π/4+a)cos(π/4+b)
代入数据可得sin(a+b)=56/65

Answer 3

3楼答案是对的