已知cos(π/4-a)=3/5,sin(5π/4+b)=-12/13,a∈(π/4,3π/4),b∈(0,π/4),求sina(a+b)的值.

题目

已知cos(π/4-a)=3/5,sin(5π/4+b)=-12/13,a∈(π/4,3π/4),b∈(0,π/4),求sina(a+b)的值.

答案

由a∈(π/4,3π/4)
得：π/4+a∈(π/2,π)
故：cos(π/4-a)=sin(π/4+a)=4/5
cos(π/4+a)=-3/5
由b∈(0,π/4)
得：π/4+b∈(π/4,π/2)
又有：sin(5π/4+b)=-sin(π/4+b)=-12/13
故：sin(π/4+b)=12/13
cos(π/4+b)=√[1-(-12/13)²]=5/13
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)
=4/5*(12/13)-(-3/5)*(5/13)
=12/13+3/13
=15/13