cos(a-b⼀2)=-1⼀9 ,sin(a⼀2-b)=2⼀3且a∈(π⼀2,π) b∈(0,π⼀2),求sin((a+b)⼀2)的值

2024年11月22日 20:02
有2个网友回答
网友(1):

(a+b)/2=(a-b/2)-(a/2-b)
所枣誉信以,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,凳轮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/虚祥27

网友(2):

a∈(π/2,π)
-b∈(-π/2,0)
-b/2∈(-π/4,0)
a-b/2∈(瞎备雹磨帆π/4,π)
cos(a-b/2)=-1/9
sin(a-b/2)=4√5/9

a∈(π/2,π)
a/滚塌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