已知sinx=3/5,x∈(π/2,π),求【sin(x+y)+sin(x-y)】/【cos（x+y）+cos（x-y）】的值
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x∈(π/2,π), cosx<0
sinx=3/5 cosx=-4/5
[sin(x+y)+sin(x-y)]/[cos(x+y)+cos(x-y)]
=(sinxcosy+cosxsiny+sinxcosy-cosxsiny)/(cosxcosy-sinxsiny+cosxcosy+sinxsiny)
=(sinxcosy)/(cosxcosy)
=sinx/(cosx)
=(3/5)/(-4/5)
=-3/4
sinx=3/5 cosx=-4/5
[sin(x+y)+sin(x-y)]/[cos(x+y)+cos(x-y)]
=(sinxcosy+cosxsiny+sinxcosy-cosxsiny)/(cosxcosy-sinxsiny+cosxcosy+sinxsiny)
=(sinxcosy)/(cosxcosy)
=sinx/(cosx)
=(3/5)/(-4/5)
=-3/4
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