题目

已知cos(α-β)=-4/5,sin(α+β)=-3/5,π/2<α-β<π,3π/2<α+β<2π,求β的值

提问时间：2021-04-22

答案

sin（α+β）=-3/5,cos（α-β）=-4/5,3π/2<α+β<2Π,Π/2<α-β<Π
π/2<β<3π/2
cos(a+β)=4/5
sin(a-β)=3/5
cos2β
=cos[(a+β)-(a-β)]
=cos(a+β)cos(a-β)+sin(a+β)sin(a-β)
=4/5*(-4/5)+(-3/5)*3/5
=-16/25-9/25
=-1
2β=π或2β=-π
β=π/2或β=-π/2