已知sin（a+2b）=4/5,cos(a+b)=12/13(a+2b,a+b）属于（0~2π）求sinb

(a+2b,a+b）属于（0~2π）
sin(a+2b)=4/5,cos(a+2b)=±根号(1-sin^2(a+2b))=±3/5
cos(a+b)=12/13 ,sin(a+b)=±根号(1-cos^2(a+b))=±5/13
sin(a+2b)cos(a+b)=4/5 * 12/13 = 48/65
cos(a+2b)sin(a+b)=(±3/5) * (±5/13)=15/65,或-15/65
sinb=sin{(a+2b)-(a+b)}
=sin(a+2b)cos(a+b)-cos(a+2b)sin(a+b)
=48/65-15/65=33/65
或：
=48/65-（-15/65）=63/65

(a+2b,a+b）属于（0~2π）；sin（a+2b）=4/5，cos(a+2b)=3/5或-3/5;
cos(a+b)=12/13 ,sin(a+b)=5/13或-5/13;
sinb=sin[(a+2b)-(a+b)]=sin(a+2b)*cos(a+b)-sin(a+b)*cos(a+2b)
=63/65或33/65