设0＜α＜β＜π/2,且cos(α+β)=-5/13,cos(α-β)=3/5,则cos2β=?
问题描述：
设0＜α＜β＜π/2,且cos(α+β)=-5/13,cos(α-β)=3/5,则cos2β=?
设0＜α＜β＜π/2,且cos(α+β)=-5/13,cos(α-β)=3/5,则cos2β=?
答
sin(a b)=12/13
sin(a-b)=-4/5
cos2b=cos[(a b)-(a-b)]=cos(a b)cos(a-b) sin(a b)sin(a-b)=-15/65-48/65=-63/65