Question

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)的值.……

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)的值.
需要解和过程，最重要的是分析。

Answer 1

解:cos(β-α)=cos(α-β)=-cos(π/3+α+2π/3-β)=-{cos[(π/3)+α]*cos(2π/3-β)-sin[(π/3)+α]*sin(2π/3-β)}
因为0＜α＜π/2＜β＜π 所以 sin[(π/3)+α]=2/5 cos(2π/3-β)=12/13
所以 cos(β-α)=-[(-3/5)*(12/13)-(2/5)*(5/13)]=46/65

Answer 2

∵0＜a＜π/2＜b＜π
∴-π/3＜(2π/3)-b＜π/6
结合题设
cos[(π/3)+a]=-3/5＜0
sin[(2π/3)-b]=5/13＞0
可知:π/2＜(π/3)+a＜π
0＜(2π/3)-b＜π/6
∴cos[(π/3)+a]=-3/5
sin[(π/3)+a]=4/5
sin[(2π/3)-b]=5/13
cos[(2π/3)-b]=12/13
[[2]]
易知
π+(a-b)=[(π/3)+a]+[(2π/3)-b]
cos[π+(a-b)]=cos[(π/3)+a]cos[(2π/3)-b]-sin[(π/3)+a]sin[(2π/3)-b]
= (-3/5)×(12/13)-(4/5)×(5/13)
=-56/65
即cos[π+(a-b)]=-56/65
又cos[π+(a-b)]=-cos(a-b)=-cos(b-a)
∴cos(b-a)=56/65

Answer 3

0＜α＜π/2
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

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

Answer 4

∵β-α= π- { [(π/3)+α] + [2π/3-β]} 角度关系分析
∵cos[(π/3)+α]=-3/5, sin(2π/3-β)=5/13 0＜α＜π/2＜β＜π
∴sin[(π/3)+α]= 4/5 cos (2π/3-β)= 12/13
∴cos(β-α)=cos π- { [(π/3)+α] + [2π/3-β]}
= - cos { [(π/3)+α] + [2π/3-β]} 诱导公式
= - cos [(π/3)+α] *cos [2π/3-β] + sin [(π/3)+α]*sin [2π/3-β]}
= 3/5 * 12/13 +4/5 * 5 /13 =56/65

Answer 5

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