Question

已知α，β∈（3π/4，π),cos（α+β)=4/5,sin(β-π/4)=12/13,求cos(α+π/4)

Answer 1

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

Answer 2

cos（α+β)=cos（(α+π/4)+(β-π/4))=cos(α+π/4)cos（β-π/4)-SIN(α+π/4)SIN（β-π/4)=4/5
sin（α+β)=sin（(α+π/4)+(β-π/4))=sin(α+π/4)cos（β-π/4)+cos(α+π/4)SIN（β-π/4)=-3/5
(IV象限)
COS(β-π/4)=-5/13(II象限)
cos(α+π/4)(-5/13)-SIN(α+π/4)(12/13)=4/5
sin(α+π/4)(-5/13)+cos(α+π/4)(12/13)=-3/5
cos(α+π/4)=-56/65