cos(a-b/2)=-1/9， sin(a/2-b)=2/3 且π/2<a<π ，0<b<π/2 求cos （(a+b)/2）的值 求详细过程~谢谢~

Question

如题所述

Answer 1

π/2<a<π ，0<b<π/2，
故π/4< a/2 <π/2 ，0<b/2 <π/4
所以
π/4 <a- b/2<π
-π/4< a/2-b <π/2

故sin(a-b/2) >0，cos(a/2 -b) >0
所以
sin(a-b/2)= √[1- cos²(a-b/2)]= √80 /9 =4√5 /9
而cos(a/2 -b)= √[1- sin²(a/2 -b)]= √5 /3

故
cos [(a+b)/2]
=cos [(a-b/2) - (a/2-b)]
=cos(a-b/2)*cos(a/2-b) + sin(a-b/2)*sin(a/2-b)
= -1/9 * √5 /3 + 4√5 /9 * 2/3
= 7√5 /27