Question

已知α，β∈(3π/4, π)，sin(α+β)=－3/5 sin(β-π/4)=12/13，则cos(α+π/4)= 解答过程

Answer 1

α，β∈(3π/4, π)，a+β∈（3π/2,2π），β-π/4∈（π/2,3π/4）
cos(α+β)=4/5 ，cos(β-π/4)=-5/13
cos(α+π/4)= =cos{（a+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

α，β∈(3π/4, π)，sin(α+β)=－3/5 sin(β-π/4)=12/13，
α+β∈(3π/2, 2π)，cos(α+β)=4/5,β-π/4∈(π/2, 3π/4)，cos(β-π/4)=-5/13
则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)+(12/13)×(-3/5)=-56/65