已知sin(a-b)cosa-cos(b-a)sina=4/5,b是第三象限的角,求sin(b+π/4)的值

Question

已知sin(a-b)cosa-cos(b-a)sina=4/5,b是第三象限的角,求sin(b+π/4)的值

Answer 1

sin(a-b)cosa-cos(b-a)sina=sin(a-b)cosa-cos(a-b)sina=sin(a-b-a)=sin(-b)=-sinb=4/5
所以,sinb=-4/5
b是第三象限角,则cosb=-3/5
所以,sin(b+π/4)=sinbcos(π/4)+cosbsin(π/4)
=(-4/5)(√2/2)+(-3/5)(√2/2)
=-7√2/10
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