Question

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

3Q.

Answer 1

1.因为(3π/4+B)-(π/4-a)=B+a+π/2
所以cos[（a+B）+π/2]=cos(a+B)*cosπ/2-sin(a+B)*sinπ/2=-sin(a+b)
所以sin(a+b)=-cos[（a+B）+π/2]=-cos[(3π/4+B)-(π/4-B)]
=-[cos(3π/4+B)*cos(π/4-a)+sin(3π/4+B)*sin(π/4-a)]
由π/4<a<3π/4,0<B<π/4，得-π/2<π/4-a<0；3π/4<3π/4+B<π
又cos(π/4-a)=3/5, sin(3π/4+B)=5/13
所以sin(π/4-a)=-4/5, cos(3π/4+B)=-12/13
因此sin(a+B)=-cos[（a+B）+π/2]=-cos[(3π/4+B)-(π/4-a)]
=-[cos(3π/4+B)*cos(π/4-a)+sin(3π/4+B)*sin(π/4-a)]
=-[3/5*(-12/13)+5/13*(-4/5)]=36/65+20/65=56/65