已知cos(π/4-a)=3/5,sin(5π/4+B)=-12/13,a∈(π/4,3/4π),B∈(0,π/4),求sin(a-B)的值
问题描述：
已知cos(π/4-a)=3/5,sin(5π/4+B)=-12/13,a∈(π/4,3/4π),B∈(0,π/4),求sin(a-B)的值
答
已知cos(π/4-a)=3/5,sin(5π/4+B)=-12/13,a∈(π/4,3/4π),B∈(0,π/4),则
π/4-a∈(-π/2,0),
sin(π/4-a)=-4/5
5π/4+B∈(5π/4,3π/2),
cos(5π/4+B)=-5/13
sin(a-B)=cos(π/2-a+B)=-cos(3π/2-a+B)=-cos[（π/4-a）+(5π/4+B)]=-[cos(π/4-a)cos(5π/4+B)-sin(π/4-a)sin(5π/4+B)]=-[（3/5）*（-5/13）-（-4/5）*（-12/13）]=63/65