Question

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

求sin（a+b）

Answer 1

π/4<a<3π/4, -π/2<π/4-a<0
cos（π/4-a）=3/5,sin（π/4-a）=-4/5
因为sin2a=cos（π/2-2a）=cos²（π/4-a）-sin²（π/4-a）=-7/25＜0
所以π<2a<3π/2，π/2<a<3π/4
cos2a=sin（π/2-2a）=2*cos（π/4-a）sin（π/4-a）=-24/25

sina=7/5根号2，cosa=-1/5根号2
0<b<π/4，0<π/4-b<π/4
sin（3π/4+b）=5/13，sin（π/4-b）=5/13，cos（π/4-b）=12/13
cos2b=sin（π/2-2b）=2*cos（π/4-b）sin（π/4-b）=120/169
sinb=7/13根号2，cosb=17/13根号2
sin（a+b）=sina*cosb+sinb*cosa
=7/5根号2*17/13根号2+7/13根号2（-1/5根号2）=56/65

Answer 2