题目

已知cos(π／3+α）=－3／5 ,sin(2π／3－β）=5／13,且0＜α＜π／2＜β＜π,求cos(β－α).

提问时间：2020-10-03

答案

0＜α＜π／2
0+π/3＜α+π/3＜π／2+π/3
π/3＜α+π/3＜5π/6
cos(π/3+α)=-3/5 ,
sin(π/3+α)=4/5 ,
π／2+π/3＜β+π/3＜π+π/3
5π/6＜β+π/3＜4π/3
cos(π/3+α)=-3/5 ,
sin(2π/3-β)=5/13
sin(π-π/3-β)=5/13
sin(π/3+β)=5/13
cos(π/3+β)=-12/13
cos(β－α)
=cos[π/3+β－(π/3+α)]
=cos(π/3+β)cos(π/3+α)+sin(π/3+β)sin(π/3+α)
=-12/13*(-3/5)+5/13*4/5
=36/65+20/65
=56/65