已知cosa-cosβ=二分之一,sina-sinβ=-三分之一,求sin(a+β)的值谢谢谢谢

cosa-cosβ=1/2,
sina-sinβ=-1/3;
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(cosa-cosβ)^2 - (sina-sinβ)^2
=cosacosa+cosβcosβ-2cosacosβ -(sinasina+sinβsinβ-2sinβsina)
=cos2a+cos2β-2cos(a+β)
=2cos(a+β)cos(a-β) - 2cos(a+β) =5/36;
cos(a+β)cos(a-β) - cos(a+β) =5/72;
[cos(a+β)][-1 + cos(a-β)]=5/72;
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(cosa-cosβ)^2 + (sina-sinβ)^2
=cosacosa+cosβcosβ-2cosacosβ +(sinasina+sinβsinβ-2sinβsina)
=cosacosa+sinasina+cosβcosβ+sinβsinβ-2cosacosβ-2sinβsina
=2 - 2cos(a-β) =13/36
1 - cos(a-β) =13/72
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cos(a+β)=(5/72)/(-13/72)=-5/13;
cosxcosx+sinxsinx=1;
sin(a+β)=±12/13