Question

若sin（3π/4+α）=5/13,cos（π/4-β）=3/5,且0＜α＜π/4＜β＜3π/4，

求cos（α+β）的值

Answer 1

sin(¾π+α)=sin[π-(¾π+α)]=sin(¼π-α)=5/13
0<α<π/4→α=¼π-arcsin(5/13)
cos(¼π-β)=cos(β-¼π)=3/5→β=¼π+arccos(3/5)
cos(α+β)=cos[½π-(arcsin(5/13)-arccos(3/5))]
=sin[arcsin(5/13)-arccos(3/5)]
=sin[arcsin(5/13)]cos[arccos(3/5)]-cos[arcsin(5/13)]sin[arccos(3/5)]
=sin[arcsin(5/13)]cos[arccos(3/5)]-cos[arccos(12/13)]sin[arcsin(4/5)]
=(5/13)(3/5)-(12/13)(4/5)
=-33/65

Answer 2

∵0<α<π/4
∴3π/4<3π/4 + α<π
∴cos(3π/4 + α)=-√1 - sin²(3π/4 + α)
=-√1 - (5/13)²=-12/13
∵π/4<β<3π/4
∴-3π/4<-β<-π/4
∴-π/2<π/4 - β<0
∴sin(π/4 - β)=-√1 - cos²(π/4 - β)
=-√1 - (3/5)²=-4/5
则cos(α+β)=sin[π/2 + (α+β)]
=sin[(3π/4 + α) - (π/4 - β)]
=sin(3π/4 + α)cos(π/4 - β) - cos(3π/4 + α) sin(π/4 - β)
=(5/13)•(3/5) - (-12/13)•(-4/5)
=15/65 - 48/65
=-33/65