Question

己知cos（兀/4一a）=3/5，sin（5兀/4十b）=-12/13，a属于（兀/4，3兀/4），b属于（O，兀/4）.求sin（a+...

己知cos（兀/4一a）=3/5，sin（5兀/4十b）=-12/13，a属于（兀/4，3兀/4），b属于（O，兀/4）.求sin（a+b）的值

Answer 1

解：
cos(π/4一a) = cos(a-π/4) = 3/5 ==> sin(a-π/4) = 4/5
sin(5π/4+b) = -sin(π/4+b) ==> sin(b+π/4) = 12/13；cos(b+π/4) =5/13
sin(a+b)=sin[(a-π/4)+(b+π/4)] = 56/65 /*两角和公式

Answer 2

由题设可得
cos(a-π/4)=3/5
sin(a-π/4)=4/5
sin(b+π/4)=12/13
cos(b+π/4)=5/13
sin(a+b)
=sin[(a-π/4)+(b+π/4)]
=(4/5)(5/13)+(3/5)(12/13)
=56/65

Answer 3

我也不会