Question

∵π/2<a<π，0<b<π/2, ∴π/4<a-b/2<π,-π/4<a/2-b<π/2. 为什么啊？？？

Answer 1

0<b<π/2
0<b/2<π/4
-π/4<-b/2<0
加上 π/2<a<π
得 π/4<a-b/2<π
π/2<a<π， π/4<a/2<π/2
0<b<π/2， -π/2<-b<0
相加： -π/4<a/2-b<π/2

即 a<x<b, c<y<d
a+c< x+y <b+d

Answer 2

∵π/2<a<π，0<b<π/2, ∵π/2<a<π，0<b<π/2,
∴0<b/2<π/4 ∴π/4<a/2<π/2 , -π/2<-b<0
∴-π/4<-b/2<0 ∴π/4+(-π/2)<a/2+(-b)<π/2+0
∴π/2+(-π/4)<a+(-b/2)<π+0 ∴-π/4<a/2-b<π/2
∴π/4<a-b/2<π

Answer 3

0<b<π/2 0<b/2<π/4 -π/4<-b/2<0 1式
π/2<a<π 2式 1式+2式 得 π/4<a-b/2<π
同理得 -π/4<a/2-b<π/2