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).

Answer 1

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

Answer 2

∵0＜α＜π/2
∴π/3＜π/3+α＜5π/6
∵cos(π/3+α)=-3/5
∴sin(π/3+α)=√[I-cos^2(π/3+α)]=√(1-9/25)=√(16/25)=4/5
∵π/2＜β＜π
∴-π＜-β＜-π/2
∴-π/3＜2π/3-β＜π/6
∵sin(2π/3-β)=5/13
∴cos(2π/3-β)=√[I-sin^(2π/3-β)]=√(1-25/169)=√(144/169)=12/13
cos(β-α)=-cos[(2π/3-β)+(π/3+α)]
=-[cos(2π/3-β)cos(π/3+α)-sin(2π/3-β)sin(π/3+α)]
=-[(12/13)*(-3/5)-(5/13)*(4/5)]
=56/65