已知:π⼀4<a<3π⼀4,0<b<π⼀4,cos(π⼀4-a)=3⼀5,sin(π⼀4+b=5⼀13,求sin(a+b)。谢谢

2024-11-13 08:40:28
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回答1:

sin(π/4-a)=-4/5 π/4cos(π/4+b)=12/13 0sin(a+b)=sin(π/4+b-(π/4-a))=sin(π/4+b)cos(π/4-a)+cos(π/4+b)sin(π/4-a)
=5/13*3/5+12/13*(-4/5)= -33/65