Question

已知(兀/4)∠a∠(3兀/4),0∠B∠(兀/4),cos(兀/4+a)=-3/5,sin(3兀/4+B)=5/13,求sin(a+B)的值

Answer 1

(兀/4)∠a∠(3兀/4),0∠B∠(兀/4),cos(兀/4+a)=-3/5 ,sin(兀/4+a)=4/5 sin(3兀/4+B)=sin(兀/4-B)=5/13,cos(兀/4-B)=12/13,sin(a+B)=sin[(兀/4+a)-(兀/4-B)]=-3/5*12/13-4/5*5/13=-56/65.

Answer 2

(兀/4)∠a∠(兀/4),0∠B∠(兀/4)，∵cos(兀/4 a)=-3/5 ,∴sin(兀/4 a)=4/5 sin(3兀/4 B)=sin(兀/4－B)=5/13,cos(兀/4-B)=12/13,sin(a B)=sin[(兀/4 a)-(兀/4-B)]=sin(兀/4 a)*cos(兀/4－B)-cos(兀/4 a)*sin(兀/4-B)=4/5*12/13－(-3/5)*5/13=63/65。

Answer 3

.... 无言