第1个回答 2012-03-05
a∈（π/2,π)
-b∈（-π/2,0)
-b/2∈（-π/4,0)
a-b/2∈（π/4,π)
cos(a-b/2)=-1/9
sin(a-b/2)=4√5/9
a∈（π/2,π)
a/2∈（π/4,π/2)
-b∈（-π/2,0)
a/2-b（-π/4,π/2)
sin(a/2-b)=2/3
cos(a/2-b)=√5/3
sin[(a+b)/2]
=sin[(a-b/2)-(a/2-b)]
=sin(a-b/2)cos(a/2-b)-cos(a-b/2)sin(a/2-b)
=4√5/9*√5/3-(-1/9)*2/3
=20/27+2/27
=22/27本回答被网友采纳
第2个回答 2012-04-14
(a+b)/2=(a-b/2)-(a/2-b)
所以，sin((a+b)/2)=sin[(a-b/2)-(a/2-b)]=sin(a-b/2)cos(a/2-b)-cos(a-b/2)sin(a/2-b)
a∈（π/2,π) b∈（0，π/2)，那么a/2∈(π/4,π/2) , b/2∈(0，π/4)
所以a-b/2∈(π/4,π)，a/2-b∈(0,π/2)
所以，sin(a-b/2)=√[1-(-1/9)²]=4√5/9，cos(a/2-b)=√[1-(2/3)²]=√5/3
故sin((a+b)/2)=sin[(a-b/2)-(a/2-b)]=sin(a-b/2)cos(a/2-b)-cos(a-b/2)sin(a/2-b)
=4√5/9 * √5/3 - (-1/9) * 2/3
=22/27