0<β<π/6<α<2π/3,cos(α+π/3)=-5/13 sin(β+2π/3)=3/5 ,求sin(α+β)?

问题描述:

0<β<π/6<α<2π/3,cos(α+π/3)=-5/13 sin(β+2π/3)=3/5 ,求sin(α+β)?

问老师。

因为pi/6所以pi/2同理2pi/3所以sin(a+pi/3)=12/13
cos(b+2pi/3)=-4/5
sin(a+pi/3+b+2pi/3)=sin(a+pi/3)cos(b+2pi/3)+cos(a+pi/3)sin(b+2pi/3)=-63/65=sin(a+b+pi)=-sin(a+b)
sin(a+b)=63/65

0<β<π/6<α<2π/3
所以:π/2