已知cos(a-b2)=-13,sin(a2-b)=14,且3π2小于a,a小于2π,π2小于b,b小于π,求COS(a+b2)的值

题目

已知cos(a-b2)=-13,sin(a2-b)=14,且3π2小于a,a小于2π,π2小于b,b小于π,求COS(a+b2)的值
是a+b的和除以2

答案

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)
∵3π2小于a,a小于2π,π2小于b,b小于π
∴π-π/4又cos(a-b2)=-13,sin(a2-b)=14
∴sin(a-b2)=-2√2/3
cos(a/2-b)=√15/4
∴COS[(a+b）/2]=-√15/12-2√2/12
=-(√15+2√2)/12