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)的值答案是-56/65【要过程】
问题描述：
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
答案是-56/65【要过程】
答
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
答
π/4＜α＜3π/4,-π/2＜π/4-α＜0cos(π/4-α)=3/5sin(π/4-α)=-4/50＜β＜π/4π/2＜3π/4-β＜3π/4sin(3π/4-β)=5/13cos(3π/4-β)=-12/13cos(α+β)=-cos[π-(α+β)]=-cos[(π/4-α)+(3π/4-β)]=-[cos[(π...