å·²ç¥sin(α+Ï/3)+sinα=-(4/5)â3ï¼-Ï/2<α<0ï¼åcos(α+2Ï/3)=?
解ï¼sin(α+Ï/3)+sinα=2sin(α+Ï/6)cos(Ï/6)=(â3)sin(α+Ï/6)=-(4/5)â3
â´sin(α+Ï/6)=-4/5
cos(α+2Ï/3)=cos[Ï/2+(α+Ï/6)]=-sin(α+Ï/6)=4/5ï¼æ
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