Question

已知π/4<a<3π/4,0<β<π/4,cos(π/4-a)=3/5,sin(3π/4-β)=-5/13,求sin(a+β)的值.

Answer 1

sin(3π/4-β)=-5/13? 应为sin(3π/4-β)=5/13,sin(3π/4-β)=sin(π/4+β)>0
π/4<a<3π/4, 0<a-π/4<π/2
0<β<π/4, π/4<β+π/4<π/2
cos(π/4-a)=3/5=cos(a-π/4)
sin(a-π/4)=4/5
sin(3π/4-β)=5/13=sin[π-(3π/4-β)]=sin(β+π/4)
cos(β+π/4)=12/13
sin(a+β)=sin(a-π/4)cos(β+π/4)+cos(a-π/4)sin(β+π/4)=(4/5)*(12/13)+(3/5)(5/13)=63/65