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π/4<a<3π/4, 0<b<π/4,
则, π/2<π/4+a<π,
3π/4<3π/4+b<π,
cos(π/4+a)=-3/5,
则,sin(π/4+a)=4/5
sin(3π/4+b)=5/13
则,cos(3π/4+b)=-12/13
sin(a+b)
=-sin(π+a+b)
=-sin[(π/4+a)+(3π/4+b)]
=-[sin(π/4+a)cos(3π/4+b)+cos(π/4+a)sin(3π/4+b)]
=-[4/5×(-12/13)+(-3/5)×5/13]
=63/65
则, π/2<π/4+a<π,
3π/4<3π/4+b<π,
cos(π/4+a)=-3/5,
则,sin(π/4+a)=4/5
sin(3π/4+b)=5/13
则,cos(3π/4+b)=-12/13
sin(a+b)
=-sin(π+a+b)
=-sin[(π/4+a)+(3π/4+b)]
=-[sin(π/4+a)cos(3π/4+b)+cos(π/4+a)sin(3π/4+b)]
=-[4/5×(-12/13)+(-3/5)×5/13]
=63/65
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