Question

a∈(π/4,3π/4),β∈(0,π/4),cos(a-π/4)=3/5,sin(3π/4+β0=5/13,求sin(a+β)

Answer 1

a∈(π/4,3π/4),则 (a-π/4)∈(0,π/2) 所以 sin(a-π/4)>0
β∈(0,π/4) 则 (3π/4+β)∈(3π/4,π) 所以cos (3π/4+β)<0
由sin²(a-π/4)+cos²(a-π/4)=1 cos(a-π/4)=3/5, (勾股定理 3 4 5)
解得 sin(a-π/4)=4/5
同理解得cos (3π/4+β)=-12/13
cos[(a-π/4)+(3π/4+β)] 即 cos(a+β+π/2)
= cos(a-π/4)cos(3π/4+β)-sin(a-π/4)sin(3π/4+β)
=(3/5)×(-12/13)-4/5×5/13
=-56/65
即 cos(a+β+π/2)=-56/65
所以 cos(a+β-π/2)=56/65
sin(a+β)
=cos[π/2-(a+β)]
= cos(a+β-π/2)
=56/65