Question

已知0＜β＜π/4,π/4＜α＜3π/4,cos(π/4-α)=3/5,sin(3π/4-β)=5/13,求cos(α+β)的值

已知0＜β＜π/4,π/4＜α＜3π/4,cos(π/4-α)=3/5,sin(3π/4-β)=5/13,求cos(α+β)的值

要过程。谢谢。
答案是-56/65【要过程】

Answer 1

π/4＜α＜3π/4,
-π/2＜π/4-α＜0
cos(π/4-α)=3/5
sin(π/4-α)=-4/5
0＜β＜π/4
π/2＜3π/4-β＜3π/4
sin(3π/4-β)=5/13
cos(3π/4-β)=-12/13
cos(α+β)
=-cos[π-(α+β)]
=-cos[(π/4-α)+(3π/4-β)]
=-[cos[(π/4-α)*cos(3π/4-β)-sin(π/4-α)*sin(3π/4-β)]
=-[3/5*(-12/13)-(-4/5)*5/13]
=-[-16/65]
=16/65

Answer 2

cos(π/4-α)=3/5 sin(π/4-α)=-4/5
sin(3π/4-β)=sin[(π/4-β)+π/2]=cos(π/4-β)=5/13 sin(π/4-β)=12/13
cos(α+β)=sin[π/2-(α+β)]=sin[(π/4-α)+(π/4-β)]
=sin(π/4-α)cos(π/4-β)+cos(π/4-α)sin(π/4-β)
=-4/5*5/13+3/5*12/13
=16/65