Question

已知0<a<π/2<b<π,又sina=3/5，cos（a+b=-4/5），则sinb等于多少

答案是24/25
要过程

Answer 1

0<a<π/2<b<π,
π/2<a+b<3π/2
cos(a+b)=-4/5 ,sina=3/5
sin(a+b)=3/5
cosa=4/5

π/2<b<π
所以sinb>0

sinb=sin(a+b-a)
=sin(a+b)cosa-sina*cos(a+b)
=3/5*4/5-3/5*(-4/5)
=12/25+12/25
=24/25

Answer 2

0<a<π/2<b<π,
π/2<a+b<3π/2
cos(a+b)=-4/5 ,sina=3/5
则可求得：
sin(a+b)=-3/5或3/5，cosa=4/5

sinb=sin(a+b-a)
=sin(a+b)cosa-sina*cos(a+b)
=-3/5*4/5-3/5*(-4/5)
=0（因为π/2<b<π，故舍去此值）
或
sinb=sin(a+b-a)
=sin(a+b)cosa-sina*cos(a+b)
=3/5*4/5-3/5*(-4/5)
=24/25

所以sinb=24/25

Answer 3

cos(a+b)=cosa*cosb - sina*sinb = -4/5
sin(a+b)=sina*cosb + cosa*sinb = 3/5
sina = 3/5, cosa = 4/5
sinb = 24/25

Answer 4

你题目写错了吧
sina=3/5??

Answer 5

cos(a+b)=cosa*cosb - sina*sinb = -4/5
sin(a+b)=sina*cosb + cosa*sinb = 3/5
sina = 3/5, cosa = 4/5
sinb = 24/25